Solution manual of Thermodynamics-Ch.16

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16-1

Chapter 16 CHEMICAL AND PHASE EQUILIBRIUM

The Kp and Equilibrium Composition of Ideal Gases

16-1C Because when a reacting system involves heat transfer, the increase-in-entropy principle relation requires a knowledge of heat transfer between the system and its surroundings, which is impractical. The equilibrium criteria can be expressed in terms of the properties alone when the Gibbs function is used.

16-2C No, the wooden table is NOT in chemical equilibrium with the air. With proper catalyst, it will reach with the oxygen in the air and burn.

16-3C They are

ν

νν

νν

ν

ν ΔΔ−

⎟⎟⎠

⎞⎜⎜⎝

⎛===

total

/)(* and ,N

PNN

NNKeK

PP

PPK

BA

DCu

BA

DC

BA

DCp

TRTGpv

BA

Dv

Cp

where .BADC ννννν −−+=Δ The first relation is useful in partial pressure calculations, the second in determining the Kp from gibbs functions, and the last one in equilibrium composition calculations.

16-4C (a) This reaction is the reverse of the known CO reaction. The equilibrium constant is then

1/ KP

(b) This reaction is the reverse of the known CO reaction at a different pressure. Since pressure has no effect on the equilibrium constant,

1/ KP

(c) This reaction is the same as the known CO reaction multiplied by 2. The quilibirium constant is then

2PK

(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,

2PK


16-2

16-5C (a) This reaction is the reverse of the known H2O reaction. The equilibrium constant is then

1/ KP

(b) This reaction is the reverse of the known H2O reaction at a different pressure. Since pressure has no effect on the equilibrium constant,

1/ KP

(c) This reaction is the same as the known H2O reaction multiplied by 3. The quilibirium constant is then

3PK

(d) This is the same as reaction (c) occurring at a different pressure. Since pressure has no effect on the equilibrium constant,

3PK

16-6C (a) No, because Kp depends on temperature only.

(b) In general, the total mixture pressure affects the mixture composition. The equilibrium constant for the reaction N O 2NO2 2+ ⇔ can be expressed as

)(

totalON

NO2O2NNO

2O

2

2

2

NO ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK

Np

The value of the exponent in this case is 2-1-1 = 0. Therefore, changing the total mixture pressure will have no effect on the number of moles of N2, O2 and NO.

16-7C (a) The equilibrium constant for the reaction 2221 COOCO ⇔+ can be expressed as

)(

totalOCO

CO2OCO2CO

2O

2

CO

2CO

2

ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

Judging from the values in Table A-28, the Kp value for this reaction decreases as temperature increases. That is, the indicated reaction will be less complete at higher temperatures. Therefore, the number of moles of CO2 will decrease and the number moles of CO and O2 will increase as the temperature increases.

(b) The value of the exponent in this case is 1-1-0.5=-0.5, which is negative. Thus as the pressure increases, the term in the brackets will decrease. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.


16-3

16-8C (a) The equilibrium constant for the reaction N 2N2 ⇔ can be expressed as

)(

totalN

N2NN

2N

2

N νν

ν

ν −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK p

Judging from the values in Table A-28, the Kp value for this reaction increases as the temperature increases. That is, the indicated reaction will be more complete at higher temperatures. Therefore, the number of moles of N will increase and the number moles of N2 will decrease as the temperature increases.

(b) The value of the exponent in this case is 2-1 = 1, which is positive. Thus as the pressure increases, the term in the brackets also increases. The value of Kp depends on temperature only, and therefore it will not change with pressure. Then to keep the equation balanced, the number of moles of the products (N) must decrease, and the number of moles of the reactants (N2) must increase.

16-9C The equilibrium constant for the reaction 2221 COOCO ⇔+ can be expressed as

)(

totalOCO

CO2OCO2CO

2O

2

CO

2CO

2

ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

Adding more N2 (an inert gas) at constant temperature and pressure will increase Ntotal but will have no direct effect on other terms. Then to keep the equation balanced, the number of moles of the products (CO2) must increase, and the number of moles of the reactants (CO, O2) must decrease.

16-10C The values of the equilibrium constants for each dissociation reaction at 3000 K are, from Table A-28,

22.359)-han (greater t 685.3ln2HH

359.22ln2NN

2

2

−=⇔⇔

−=⇔⇔

p

p

K

K

Thus H2 is more likely to dissociate than N2.


16-4

16-11 The mole fractions of the constituents of an ideal gas mixture is given. The Gibbs function of the CO in this mixture at the given mixture pressure and temperature is to be determined.

Analysis From Tables A-21 and A-26, at 1 atm pressure,

[ ]

kJ/kmol 837,244)543.1972988669()162.227800844,23(150,137

)()(atm) 1 K, (800* oo

−=×−−×−+−=

−Δ+= TsTThgg f

The partial pressure of CO is

atm 3atm) 10)(30.0(COCO === PyP

The Gibbs function of CO at 800 K and 3 atm is

kJ/kmol 237,530−=+−=+=

atm) K)ln(3 00kJ/kmol)(8 314.8(kJ/kmol 837,244lnatm) 1 K, 800(*atm) 3 K, 800( COPTRgg u

16-12 The partial pressures of the constituents of an ideal gas mixture is given. The Gibbs function of the nitrogen in this mixture at the given mixture pressure and temperature is to be determined.

Analysis The partial pressure of nitrogen is

atm 283.1)325.101/130(kPa 130N2 ===P

The Gibbs function of nitrogen at 298 K and 3 atm is

kJ/kmol 617.4=+=

+=atm) K)ln(1.283 98kJ/kmol)(2 314.8(0

lnatm) 1 K, 298(*atm) 3 K, 800( N2PTRgg u

N2 ,CO2, NO

PN2 = 130 kPa 298 K

10% CO2 60% H2O 30% CO 10 atm 800 K


16-5

16-13E The equilibrium constant of the reaction 221

22 OHOH +⇔ is to be determined using Gibbs function.

Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

K e K G T R TpG T R T

p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(* H2OH2OO2O2H2H2 TgTgTgTG ∗∗∗ −+=Δ ννν

At 1440 R,

Btu/lbmol 632,87)428.53144042584.933,11040,104(1

)326.5614401.37250.532,100(5.0)079.3814403.36409.99560(1

])[(

])[(

])[(

)()()(

)()()()(*

H2O5371440H2O

O25371440O2

H25371440H2

H2OH2OO2O2H2H2

H2OH2OO2O2H2H2

=×−−+−×−

×−−+×+×−−+×=

−−+−

−−++

−−+=

−−−+−=

−+=Δ ∗∗∗

sThhh

sThhh

sThhh

sThsThsTh

TgTgTgTG

f

f

f

ν

ν

ν

ννν

ννν

Substituting,

30.64−−= =R)] R)(1440Btu/lbmol. /[(1.986Btu/lbmol) 632,87(ln pK

or

)ioninterpolatby 97.34ln :28-A (Table −=×= −pp KK 14104.93

At 3960 R,

Btu/lbmol 436,53)402.6439604258989,39040,104(1

)032.6539601.3725441,320(5.0)765.4539603.36405.370,290(1

])[(

])[(

])[(

)()()(

)()()()(*

H2O5373960H2O

O25373960O2

H25373960H2

H2OH2OOO2H2H2

H2OH2OO2O2H2H2

2

=×−−+−×−

×−−+×+×−−+×=

−−+−

−−++

−−+=

−−−+−=

−+=Δ ∗∗∗

sThhh

sThhh

sThhh

sThsThsTh

TgTgTgTG

f

f

f

ν

ν

ν

ννν

ννν

Substituting,

6.79−−= =R)] R)(3960Btu/lbmol. /[(1.986Btu/lbmol) 436,53(ln pK

or )768.6ln :28-A (Table −=×= −pp KK 3101.125

H2O ↔ H2 + ½O2

1440 R


16-6

16-14 The reaction 222 OH2O2H +⇔ is considered. The mole fractions of the hydrogen and oxygen produced when this reaction occurs at 4000 K and 10 kPa are to be determined.

Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixture are ideal gases.

Analysis The stoichiometric and actual reactions in this case are

Stoichiometric: )1 and ,2 ,2 (thus OH2O2H O2H2H2O222 ===+⇔ ννν

Actual: 43421321

products22

react.22 O+HOHO2H zyx +⎯→⎯

H balance: xyyx −=⎯→⎯+= 2224

O balance: xzx 5.01z22 −=⎯→⎯+=

Total number of moles: xzyxN 5.03total −=++=

The equilibrium constant relation can be expressed as

)(

totalH2O

O2H2H2OO2H2

H2O

O2H2 ννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

From Table A-28, K 4000at 542.0ln −=pK . Since the stoichiometric reaction being considered is double this reaction,

3382.0)542.02exp( =×−=pK

Substituting,

212

2

2

5.03325.101/10)5.01()2(3382.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.4446

Then,

y = 2 − x = 1.555

z = 1 − 0.5x = 0.7777

Therefore, the equilibrium composition of the mixture at 4000 K and 10 kPa is

222 O 0.7777H 1.555+OH 0.4446 +

The mole fractions of hydrogen and oxygen produced are

0.280

0.560

===

==×−

==

778.27777.0

778.2555.1

4446.05.03555.1

total

O2O2

total

H2H2

NN

y

NN

y

2H2O 4000 K 10 kPa


16-7

16-15 The reaction 222 OH2O2H +⇔ is considered. The mole fractions of hydrogen gas produced is to be determined at 100 kPa and compared to that at 10 kPa.

Assumptions 1 The equilibrium composition consists of H2O, H2, and O2. 2 The constituents of the mixture are ideal gases.



Actual: 43421321

products22

react.22 O+HOHO2H zyx +⎯→⎯



Total number of moles: xzyxN 5.03total −=++=


)(

totalH2O

O2H2H2OO2H2

H2O

O2H2 ννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p


3382.0)542.02exp( =×−=pK

Substituting,

212

2

2

5.03325.101/100)5.01()2(3382.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.8870

Then,

y = 2 − x = 1.113

z = 1 − 0.5x = 0.5565


222 O 0.5565H 1.113+OH 0.8870 +

That is, there are 1.113 kmol of hydrogen gas. The mole number of hydrogen at 10 kPa reaction pressure was obtained in the previous problem to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has decreased.

2H2O 4000 K 100 kPa


16-8

16-16 The reaction 222 OH2O2H +⇔ is considered. The mole number of hydrogen gas produced is to be determined if inert nitrogen is mixed with water vapor is to be determined and compared to the case with no inert nitrogen.

Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases.



Actual: 32143421321

inert2

products22

react.222 0.5NO+HOH0.5NO2H ++⎯→⎯+ zyx



Total number of moles: xzyxN 5.05.35.0total −=+++=


)(

totalH2O

O2H2H2OO2H2

H2O

O2H2 ννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p


3382.0)542.02exp( =×−=pK

Substituting,

212

2

2

5.05.3325.101/10)5.01()2(3382.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.4187

Then,

y = 2 − x = 1.581

z = 1 − 0.5x = 0.7907


222 O 0.7907H 1.581+OH 0.4187 +

That is, there are 1.581 kmol of hydrogen gas. The mole number of hydrogen without inert nitrogen case was obtained in Prob. 16-14 to be 1.555 kmol. Therefore, the amount of hydrogen gas produced has increased.

2H2O, 0.5N2 4000 K 10 kPa


16-9

16-17E The temperature at which 15 percent of diatomic oxygen dissociates into monatomic oxygen at two pressures is to be determined.

Assumptions 1 The equilibrium composition consists of O2 and O. 2 The constituents of the mixture are ideal gases.

Analysis (a) The stoichiometric and actual reactions can be written as

Stoichiometric: )2 and 1 (thus O2O OO22 ==⇔ νν

Actual: {prod.react.

22 O3.0O85.0O +⇔43421

The equilibrium constant Kp can be determined from

01880.03.085.0

696.14/385.03.0 122

totalO2

OO2O

O2

O

=⎟⎠⎞

⎜⎝⎛

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−νν

ν

ν

NP

N

NK p

and

974.3ln −=pK

From Table A-28, the temperature corresponding to this lnKp value is

T = 3060 K = 5508 R

(b) At 100 psia,

6265.03.085.0

696.14/10085.03.0 122

totalO2

OO2O

O2

O

=⎟⎠⎞

⎜⎝⎛

+=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−νν

ν

ν

NP

NN

K p

4676.0ln −=pK

From Table A-28, the temperature corresponding to this lnKp value is

T = 3701 K = 6662 R

O2 ↔ 2O 15 % 3 psia


16-10

16-18 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined.

Assumptions 1 The equilibrium composition consists of CO2, CO, and O. 2 The constituents of the mixture are ideal gases.

Analysis For the stoichiometric reaction OCOCO 221

2 +⇔ , from Table A-28, at 2500 K

331.3ln −=pK

For the oxygen dissociation reaction O0.5O 2 ⇔ , from Table A-28, at 2500 K,

255.42/509.8ln −=−=pK

For the desired stoichiometric reaction )1 and 1 ,1 (thus OCOCO OCOCO22 ===+⇔ ννν ,

586.7255.4331.3ln −=−−=pK

and

0005075.0)586.7exp( =−=pK

Actual: 43421321

productsreact.22 O+COCOCO zyx +⎯→⎯

C balance: xyyx −=⎯→⎯+= 11

O balance: xzyx −=⎯→⎯++= 1z22

Total number of moles: xzyxN −=++= 2total


CO2OCO

CO2

OCO

totalCO2

OCOννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

Substituting,

111

21)1)(1(0005075.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.9775

Then,

y = 1 − x = 0.0225

z = 1 − x = 0.0225

Therefore, the equilibrium composition of the mixture at 2500 K and 1 atm is

O 0.0225CO 0.0225+CO 0.9775 2 +

CO2 2500 K 1 atm


16-11

16-19 The dissociation reaction CO2 ⇔ CO + O is considered. The composition of the products at given pressure and temperature is to be determined when nitrogen is added to carbon dioxide.

Assumptions 1 The equilibrium composition consists of CO2, CO, O, and N2. 2 The constituents of the mixture are ideal gases.

Analysis For the stoichiometric reaction OCOCO 221

2 +⇔ , from Table A-28, at 2500 K

331.3ln −=pK

For the oxygen dissociation reaction O0.5O 2 ⇔ , from Table A-28, at 2500 K,

255.42/509.8ln −=−=pK

For the desired stoichiometric reaction )1 and 1 ,1 (thus OCOCO OCOCO22 ===+⇔ ννν ,

586.7255.4331.3ln −=−−=pK

and

0005075.0)586.7exp( =−=pK

Actual: {inert

2productsreact.

222 N3O+COCON3CO ++⎯→⎯+43421321zyx


O balance: xzyx −=⎯→⎯++= 1z22

Total number of moles: xzyxN −=+++= 53total


CO2OCO

CO2

OCO

totalCO2

OCOννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

Substituting,

111

51)1)(1(0005075.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.9557

Then,

y = 1 − x = 0.0443

z = 1 − x = 0.0443


22 3NO 0.0443CO 0.0443+CO 0.9557 ++

CO2, 3N2 2500 K 1 atm


16-12

16-20 It is to be shown that as long as the extent of the reaction, α, for the disassociation reaction X2 ⇔ 2X

is smaller than one, α is given by P

P

KK+

=4

α

Assumptions The reaction occurs at the reference temperature.

Analysis The stoichiometric and actual reactions can be written as

Stoichiometric: )2 and 1 (thus X2X XX22 ==⇔ νν

Actual: {prod.react.

22 2X)1(X Xαα +−⇔43421

The equilibrium constant Kp is given by

)1)(1(

41

1)1(

)2( 2122

totalX2

XX2X

X2

X

ααα

ααα

νν

ν

ν

+−=⎟

⎠⎞

⎜⎝⎛

+−=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

NP

N

NK p

Solving this expression for α gives

P

P

KK+

=4

α


16-13

16-21 A gaseous mixture consisting of methane and nitrogen is heated. The equilibrium composition (by mole fraction) of the resulting mixture is to be determined. Assumptions 1 The equilibrium composition consists of CH4, C, H2, and N2. 2 The constituents of the mixture are ideal gases. Analysis The stoichiometric and actual reactions in this case are Stoichiometric: )2 and ,1 ,1 (thus 2HC CH H2CCH424 ===+⇔ ννν

Actual: {inert

2products

2react.

424 NH+CCHNCH ++⎯→⎯+43421321zyx


H balance: xzx 22z244 −=⎯→⎯+=

Total number of moles: xzyxN 241total −=+++=


H2CCH4

H2C

CH4

totalH2C

CH4ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From the problem statement at 1000 K, 328.2ln =pK . Then,

257.10)328.2exp( ==pK

For the reverse reaction that we consider, 09749.0257.10/1 ==pK

Substituting,

211

2 241

)22)(1(09749.0

−−

⎟⎠⎞

⎜⎝⎛

−−−=

xxxx

Solving for x, x = 0.02325 Then, y = 1 − x = 0.9768 z = 2 − 2x = 1.9535 Therefore, the equilibrium composition of the mixture at 1000 K and 1 atm is 224 N 1H 9535.1C 0.9768+CH 0.02325 ++

The mole fractions are

0.2529

0.4941

0.2470

0.0059

===

===

===

==×−

==

954.31954.3

9535.1954.39768.0

954.302325.0

02325.02402325.0

total

N2N2

total

H2H2

total

CC

total

CH4CH4

NN

y

NN

y

NN

y

NN

y

CH4, N2 1000 K 1 atm


16-14

16-22 The reaction N2 + O2 ⇔ 2NO is considered. The equilibrium mole fraction of NO 1000 K and 1 atm is to be determined.

Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases.


Stoichiometric: )2 and ,1 ,1 (thus NO2ON NOO2N222 ===⇔+ ννν

Actual: {productsreact.

2222 NOONON zyx ++⎯→⎯+43421

N balance: xzzx 2222 −=⎯→⎯+=

O balance: xyzy =⎯→⎯+= 22

Total number of moles: 2total =++= zyxN


)(

totalO2N2

NOO2N2NO

O2N2

NO ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, at 1000 K, 388.9ln −=pK . Since the stoichiometric reaction being considered is double this reaction,

910009.7)388.92exp( −×=×−=pK

Substituting,

112

2

29

21)22(10009.7

−−− ⎟

⎠⎞

⎜⎝⎛−

=×x

x

Solving for x,

x = 0.999958

Then,

y = x = 0.999958

z = 2 − 2x = 8.4×10-5


NO 104.8O 999958.0N 0.999958 522

−×++

The mole fraction of NO is then

million)per parts (42 2104.8 5

total

NONO

5104.2 −−

×=×

==NN

y

N2, O2 1000 K 1 atm


16-15

16-23 Oxygen is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases.


Analysis (a) Obtaining oxygen properties from table A-19, an energy balance gives

kJ/kmol 50,989=−=

−=

Δ=−

6203192,5712in

energies etc. potential, kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

outin

uuq

EEE4342143421

(b) The stoichiometric and actual reactions in this case are


Actual: { {productsreact.

22 OOO yx +⎯→⎯

O balance: xyyx 2222 −=⎯→⎯+=

Total number of moles: xyxN −=+= 2total


O2O

O2

O

totalO2

Oνν

ν

ν −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK p

From Table A-28, at 2200 K, 827.11ln −=pK . Then,

610305.7)827.11exp( −×=−=pK

Substituting,

122

6

21)22(10305.7

−− ⎟

⎠⎞

⎜⎝⎛

−−

=×xx

x

Solving for x,

x = 0.99865

Then,

y = 2 − 2x = 0.0027


O 0027.0O 0.99865 2 +

Hence, the oxygen ions are negligible and the result is same as that in part (a),

kJ/kmol 50,989=inq

O2

2200 K 1 atm


16-16

16-24 Air is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases.

Assumptions 1 The equilibrium composition consists of O2 and O, and N2. 2 The constituents of the mixture are ideal gases.

Analysis (a) Obtaining air properties from table A-17, an energy balance gives

kJ/kg 1660=−=

−=

Δ=−

64.2124.187212in


system


outin

uuq

EEE4342143421



Actual: { { 43421inert

2productsreact.

222 N76.3OON76.3O ++⎯→⎯+ yx

O balance: xyyx 2222 −=⎯→⎯+=

Total number of moles: xyxN −=++= 76.576.3total


O2O

O2

O

totalO2

Oνν

ν

ν −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK p

From Table A-28, at 2200 K, 827.11ln −=pK . Then,

610305.7)827.11exp( −×=−=pK

Substituting,

122

6

76.51)22(10305.7

−− ⎟

⎠⎞

⎜⎝⎛

−−

=×xx

x

Solving for x,

x = 0.99706

Then,

y = 2 − 2x = 0.00588


22 N 76.3O 00588.0O 0.99706 ++

Hence, the atomic oxygen is negligible and the result is same as that in part (a),

kJ/kg 1660=inq

O2, 3.76N2

2200 K 1 atm


16-17

16-25 The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data.



p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(*222222 OOHHOHOH TgTgTgTG ∗∗∗ −−=Δ ννν

At 25°C,

ΔG T*( ) ( , ) ( ) . ( ) ,= − − − = −1 228 590 1 0 0 5 0 228 590 kJ / kmol

Substituting,

92.26=K)] K)(298kJ/kmol (8.314kJ/kmol)/[ 590,228(ln ⋅−−=pK

or

K Kp p= × =1.12 1040 (Table A - 28: ln . )92 21

(b) At 2000 K,

kJ/kmol 556,135)655.2682000)8682881,670(5.0

)297.18820008468400,610(1)571.26420009904593,82820,241(1

])[(

])[(

])[(

)()()(

)()()()(*

22

22

22

222222

222222

O2982000O

H2982000H

OH2982000OH

OOHHOHOH

OOHHOHOH

−=×−−+×−

×−−+×−×−−+−×=

−−+−

−−+−

−−+=

−−−−−=

−−=Δ ∗∗∗

sThhh

sThhh

sThhh

sThsThsTh

TgTgTgTG

f

f

f

ν

ν

ν

ννν

ννν

Substituting,


or

)145.8ln :28-A (Table == pp KK 3471

H2 + ½O2 ↔ H2O

25ºC


16-18

16-26E The equilibrium constant of the reaction H2 + 1/2O2 ↔ H2O is listed in Table A-28 at different temperatures. The data are to be verified at two temperatures using Gibbs function data.



p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(*222222 OOHHOHOH TgTgTgTG ∗∗∗ −−=Δ ννν

At 537 R,

ΔG T*( ) ( , ) ( ) . ( ) ,= − − − = −1 98 350 1 0 05 0 98 350 Btu / lbmol

Substituting,

ln ( ,K p = − − ⋅98 350 Btu / lbmol) / [(1.986 Btu / lbmol R)(537 R)] = 92.22

or

K Kp p= × =1.12 1040 (Table A - 28: ln . )92 21

(b) At 3240 R,

Btu/lbmol 385,63)224.6332401.3725972,250(5.0)125.4432403.36407.484,230(1

)948.61324042585.204,31040,104(1

])[(

])[(

])[(

)()()(

)()()()(*

22

22

22

222222

222222

O2983240O

H2983240H

OH5373240OH

OOHHOHOH

OOHHOHOH

−=×−−+×−×−−+×−

×−−+−×=

−−+−

−−+−

−−+=

−−−−−=

−−=Δ ∗∗∗

sThhh

sThhh

sThhh

sThsThsTh

TgTgTgTG

f

f

f

ν

ν

ν

ννν

ννν

Substituting,

9.85=R)] R)(3240Btu/lbmol. /[(1.986Btu/lbmol) 385,63(ln −−=pK

or

K Kp p= × =1.90 104 (Table A - 28: ln . )9 83

H2 + ½O2 ↔ H2O

537 R


16-19

16-27 The equilibrium constant of the reaction CO + 1/2O2 ↔ CO2 at 298 K and 2000 K are to be determined, and compared with the values listed in Table A-28.



p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(* O2O2COCOCO2CO2 TgTgTgTG ∗∗∗ −−=Δ ννν

At 298 K,

kJ/kmol 210,257)0(5.0)150,137(1)360,394(1)(* −=−−−−=Δ TG

where the Gibbs functions are obtained from Table A-26. Substituting,

103.81=⋅

−−=

K) K)(298kJ/kmol (8.314kJ/kmol) 210,257(ln pK

From Table A-28: 103.76=pKln

(b) At 2000 K,

[ ] [ ] [ ]

kJ/kmol 409,110)53.268)(2000()193,59(5.0)48.258)(2000()826,53(1)00.309)(2000()128,302(1

)()()(

)()()()(*

O2O2COCOCO2CO2

O2O2COCOCO2CO2

−=−−−−−−−=

−−−−−=

−−=Δ ∗∗∗

sThsThsTh

TgTgTgTG

ννν

ννν

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa (1 atm) are obtained from EES. Substituting,

6.64=⋅

−−=

K) K)(2000kJ/kmol (8.314kJ/kmol) 409,110(ln pK

From Table A-28:

6.635=pKln

2CO2O21CO ⇔+

298 K


16-20

16-28 EES The effect of varying the percent excess air during the steady-flow combustion of hydrogen is to be studied.

Analysis The combustion equation of hydrogen with stoichiometric amount of air is

[ ] 22222 N )76.3(5.0OH3.76NO5.0H +⎯→⎯++

For the incomplete combustion with 100% excess air, the combustion equation is

[ ] 2222222 N O H OH 97.03.76NO)5.0)(1(H cbaEx +++⎯→⎯+++

The coefficients are to be determined from the mass balances

Hydrogen balance: 03.02297.02 =⎯→⎯×+×= aa

Oxygen balance: 297.025.0)1( ×+=××+ bEx

Nitrogen balance: 2276.35.0)1( ×=×××+ cEx

Solving the above equations, we find the coefficients (Ex = 1, a = 0.03 b = 0.515, c = 3.76) and write the balanced reaction equation as

[ ] 2222222 N 76.3O 515.0H 03.0OH 97.03.76NOH +++⎯→⎯++

Total moles of products at equilibrium are

275.576.3515.003.097.0tot =+++=N

The assumed equilibrium reaction is

222 O5.0HOH +⎯→←

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using


p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(* prodH2OH2OprodO2O2prodH2H2 TgTgTgTG ∗∗∗ −+=Δ ννν

and the Gibbs functions are defined as

H2OprodprodH2O

O2prodprodO2

H2prodprodH2

)()(

)()(

)()(

sThTg

sThTg

sThTg

−=

−=

−=

∗

∗

∗

The equilibrium constant is also given by

009664.097.0

)515.0)(03.0(275.51

97.0

5.05.0

1

5.015.01

=⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−+ab

NPKtot

p

and 647.4)009664.0ln(ln −==pK

The corresponding temperature is obtained solving the above equations using EES to be

K 2600=prodT

This is the temperature at which 97 percent of H2 will burn into H2O. The copy of EES solution is given next.


16-21

"Input Data from parametric table:" {PercentEx = 10} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The combustion equation of H2 with stoichiometric amount of air is H2 + 0.5(O2 + 3.76N2)=H2O +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is H2 + (1+EX)(0.5)(O2 + 3.76N2)=0.97 H2O +aH2 + bO2+cN2" "Specie balance equations give the values of a, b, and c." "H, hydrogen" 2 = 0.97*2 + a*2 "O, oxygen" (1+Ex)*0.5*2=0.97 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is H2O=H2+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each H2mponent in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_H2O=Enthalpy(H2O,T=T_prod )-T_prod *Entropy(H2O,T=T_prod ,P=101.3) g_H2=Enthalpy(H2,T=T_prod )-T_prod *Entropy(H2,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_H2+0.5*g_O2-1*g_H2O "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P)

ln Kp PercentEx [%]

Tprod [K]

-5.414 10 2440 -5.165 20 2490 -5.019 30 2520 -4.918 40 2542 -4.844 50 2557 -4.786 60 2570 -4.739 70 2580

-4.7 80 2589 -4.667 90 2596 -4.639 100 2602

10 20 30 40 50 60 70 80 90 1002425

2465

2505

2545

2585

2625

PercentEx

T pro

d


16-22

16-29 The equilibrium constant of the reaction CH4 + 2O2 ↔ CO2 + 2H2O at 25°C is to be determined.

Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using


p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()()(*22442222 OOCHCHOHOHCOCO TgTgTgTgTG ∗∗∗∗ −−+=Δ νννν

At 25°C,

ΔG T*( ) ( , ) ( , ) ( , ) ( ) ,= − + − − − − = −1 394 360 2 228 590 1 50 790 2 0 800 750 kJ / kmol

Substituting,


or

Kp = ×1.96 10140

CH4 + 2O2 ↔ CO2 + 2H2O

25°C


16-23

16-30 The equilibrium constant of the reaction CO2 ↔ CO + 1/2O2 is listed in Table A-28 at different temperatures. It is to be verified using Gibbs function data.


TRTGKeK upTRTG

pu /)(*lnor /)*( Δ−== Δ−

where )()()()(*2222 COCOOOCOCO TgTgTgTG ∗∗∗ −+=Δ ννν

At 298 K,

kJ/kmol 210,257)360,394(1)0(5.0)150,137(1)(* =−−+−=Δ TG

Substituting,

-103.81=K)] K)(298kJ/kmol (8.314kJ/kmol)/[ 210,257(ln ⋅−=pK

or )76.103ln :28-A (Table −=×= pp KK -46108.24

(b) At 1800 K,

kJ/kmol 2.240,127)884.30218009364806,88520,393(1

)701.26418008682371,600(5.0)797.25418008669191,58530,110(1

])[(

])[(

])[(

)()()(

)()()()(*

22

22

2222

2222

CO2981800CO

O2981800O

CO2981800CO

COCOOOCOCO

COCOOOCOCO

=×−−+−×−

×−−+×+×−−+−×=

−−+−

−−++

−−+=

−−−+−=

−+=Δ ∗∗∗

sThhh

sThhh

sThhh

sThsThsTh

TgTgTgTG

f

f

f

ν

ν

ν

ννν

ννν

Substituting, .5028=K)] K)(1800kJ/kmol (8.314kJ/kmol)/[ 2.240,127(ln −⋅−=pK

or )497.8ln :28-A (Table −=×= pp KK -4102.03

CO2 ↔ CO + ½O2 298 K


16-24

16-31 [Also solved by EES on enclosed CD] Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined.

Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and N2. 2 The constituents of the mixture are ideal gases.

Analysis Assuming N2 to remain as an inert gas, the stoichiometric and actual reactions can be written as

Stoichiometric: ) and ,1 ,1 (thus COO+CO 21

OCOCO2221

22===⇔ ννν

Actual: CO +1(O N CO CO + 0.515O N2 2product

2reactants inert

+ ⎯→⎯ + +376 0 97 0 03 3762 2. ) . . .124 34 1 2444 3444 124 34


481037635150030970

15150030

970 511

50

totalOCO

CO2OCO2CO

2O

2CO

2CO

2

.......

.

NP

NN

NK

.

.

)(

p

=

⎟⎠⎞

⎜⎝⎛

+++×=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

ν−ν−ν

νν

ν

From Table A-28, the temperature corresponding to this Kp value is T = 2276 K

CO + ½O2 ↔ CO2 97 % 1 atm


16-25

16-32 EES Problem 16-31 is reconsidered. The effect of varying the percent excess air during the steady-flow process from 0 to 200 percent on the temperature at which 97 percent of CO burn into CO2 is to be studied.

Analysis The problem is solved using EES, and the solution is given below.

"To solve this problem, we need to give EES a guess value for T_prop other than the default value of 1. Set the guess value of T_prod to 1000 K by selecting Variable Infromation in the Options menu. Then press F2 or click the Calculator icon." "Input Data from the diagram window:" {PercentEx = 100} Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] "The combustion equation of CO with stoichiometric amount of air is CO + 0.5(O2 + 3.76N2)=CO2 +0.5(3.76)N2" "For the incomplete combustion with 100% excess air, the combustion equation is CO + (!+EX)(0.5)(O2 + 3.76N2)=0.97 CO2 +aCO + bO2+cN2" "Specie balance equations give the values of a, b, and c." "C, Carbon" 1 = 0.97 + a "O, oxygen" 1 +(1+Ex)*0.5*2=0.97*2 + a *1 + b*2 "N, nitrogen" (1+Ex)*0.5*3.76 *2 = c*2 N_tot =0.97+a +b +c "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(a^1*b^0.5)/(0.97^1)" sqrt(P/N_tot )*a *sqrt(b )=K_P *0.97 lnK_p = ln(k_P) "Compare the value of lnK_p calculated by EES with the value of lnK_p from table A-28 in the text."


16-26

PercentEx

[%] Tprod [K]

0 2066 20 2194 40 2230 60 2250 80 2263

100 2273 120 2280 140 2285 160 2290 180 2294 200 2297

0 40 80 120 160 2002050

2100

2150

2200

2250

2300

2350

PercentEx [%]

T pro

d [K

]


16-27

16-33E Carbon monoxide is burned with 100 percent excess air. The temperature at which 97 percent of CO burn to CO2 is to be determined.




OCOCO2221

22===⇔ ννν

Actual: CO +1(O N CO CO + 0.515O N2 2product

2reactants inert

+ ⎯→⎯ + +376 0 97 0 03 3762 2. ) . . .124 34 1 2444 3444 124 34


481037635150030970

15150030

970 511

50

totalOCO

CO2OCO2CO

2O

2

CO

2CO

2

.......

.

NP

NN

NK

.

.

)(

p

=

⎟⎠⎞

⎜⎝⎛

+++×=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

ν−ν−ν

νν

ν

From Table A-28, the temperature corresponding to this Kp value is T = 2276 K = 4097 R

16-34 Hydrogen is burned with 150 percent theoretical air. The temperature at which 98 percent of H2 will burn to H2O is to be determined.

Assumptions 1 The equilibrium composition consists of H2O, H2, O2, and N2. 2 The constituents of the mixture are ideal gases.


Stoichiometric: H + O H O (thus and 212 2 H O H O2 2 22

121 1⇔ = = =ν ν ν, , )

Actual: H + 0.75(O N H O H + 0.26O N2 2 2product

2 2reactants inert

+ ⎯ →⎯ + +376 0 98 0 02 2 822 2. ) . . .1 24 34 1 2444 3444 124 34


11.19482.226.002.098.0

126.002.0

98.0 5.11

5.0

)(

totalOH

OH2O2HO2H

2O

2

2H

2

O2H

2

=

⎟⎠⎞

⎜⎝⎛

+++×=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

−

−− ννν

νν

ν

NP

NN

NK p

From Table A-28, the temperature corresponding to this Kp value is T = 2472 K.

CO + ½O2 ↔ CO2 97 % 1 atm

H2O, H2 O2, N2

H2

Air

Combustion chamber


16-28

16-35 Air is heated to a high temperature. The equilibrium composition at that temperature is to be determined.

Assumptions 1 The equilibrium composition consists of N2, O2, and NO. 2 The constituents of the mixture are ideal gases.


Stoichiometric: ) and , ,1 (thus NOO+N 21

O21

NNO221

221

22===⇔ ννν

Actual: 3.76 N + O NO N + O2 2prod.

2 2reactants

⎯→⎯ +x y z123 1 24 34

N balance: 7.52 = x + 2y or y = 3.76 - 0.5x

O balance: 2 = x + 2z or z = 1 - 0.5x

Total number of moles: Ntotal = x + y + z = x + 4.76- x = 4.76


)(

totalON

NO2O2NNO

2O

2

2N

2

NOννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, ln Kp = -3.931 at 2000 K. Thus Kp = 0.01962. Substituting,

11

5.05.0 76.42

)5.01()5.076.3(01962.0

−

⎟⎠⎞

⎜⎝⎛

−−=

xxx

Solving for x,

x = 0.0376

Then,

y = 3.76-0.5x = 3.7412

z = 1-0.5x = 0.9812


0 0376 0 9812. .NO + 3.7412N O2 2+

The equilibrium constant for the reactions O O2 ⇔ 2 (ln Kp = -14.622) and N N2 ⇔ 2 (ln Kp = -41.645) are much smaller than that of the specified reaction (ln Kp = -3.931). Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture. Also the equilibrium composition is in this case is independent of pressure since Δν = − − =1 0 5 05 0. . .

AIR 2000 K 2 atm


16-29

16-36 Hydrogen is heated to a high temperature at a constant pressure. The percentage of H2 that will dissociate into H is to be determined.

Assumptions 1 The equilibrium composition consists of H2 and H. 2 The constituents of the mixture are ideal gases.


Stoichiometric: )2 and 1 (thus 2HH HH2 2==⇔ νν

Actual: { {prod.react.

22 HHH yx +⎯→⎯

H balance: 2 = 2x + y or y = 2 - 2x

Total number of moles: Ntotal = x + y = x + 2 - 2x = 2 - x


2HH

2H

2

H

totalH

Hνν

ν

ν −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NNK p

From Table A-28, ln Kp = -2.534 at 3200 K. Thus Kp = 0.07934. Substituting,

122

28)22(07934.0

−

⎟⎠⎞

⎜⎝⎛

−−

=xx

x

Solving for x, x = 0.95

Thus the percentage of H2 which dissociates to H at 3200 K and 8 atm is

1 - 0.95 = 0.05 or 5.0%

H2 3200 K 8 atm


16-30

16-37E A mixture of CO, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined.




OCOCO2221

22===⇔ ννν

Actual: 2CO + 2O N CO CO + O N2 2 2products

2reactants

2inert

+ ⎯→⎯ + +6 6x y z123 1 24 344 :

C balance: 2 2= + ⎯→⎯ = −x y y x

O balance: 6 2 2 2 0 5= + + ⎯→⎯ = −x y z z x.

Total number of moles: N x y z xtotal = + + + = −6 10 0 5.


)(

totalOCO

CO2OCO2CO

2O

2

CO

2CO

2

ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, .465.47 Thus K. 2400 4320at 860.3ln ==== pp KRTK Substituting,

5.11

5.0 5.0103

)5.02)(2(465.47

−

⎟⎠⎞

⎜⎝⎛

−−−=

xxxx

Solving for x,

x = 1.930

Then,

y = 2 - x = 0.070

z = 2 - 0.5x = 1.035


1.930CO + 0.070CO 1.035O 6N2 2 2+ +

2 CO 2 O2 6 N2

4320 R 3 atm


16-31

16-38 A mixture of N2, O2, and Ar is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined.

Assumptions 1 The equilibrium composition consists of N2, O2, Ar, and NO. 2 The constituents of the mixture are ideal gases.


Stoichiometric: ) and , ,1 (thus NOO+N 21

O21

NNO221

221

22===⇔ ννν

Actual: 3 01 01N + O Ar NO N + O Ar2 2prod.

2 2reactants inert

+ ⎯→⎯ + +. .x y z123 1 24 34 123

N balance: 6 2 3 0 5= + ⎯→⎯ = −x y y x.

O balance: 2 2 1 0 5= + ⎯→⎯ = −x z z x.

Total number of moles: N x y ztotal = + + + =01 41. .

The equilibrium constant relation becomes,

50501

total5050

total

22

2

2

2

2

..

..

)νν(ν

νO

νN

νNO

p NP

zyx

NP

NN

NK

ONNO

ON

NO−−−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

From Table A-28, .04885.0 Thus K. 2400at 019.3ln =−= pp KK Substituting,

1)5.01()5.03(

04885.0 5.05.0 ×−−

=xx

x

Solving for x,

x = 0.0823

Then,

y = 3 - 0.5x = 2.9589

z = 1 - 0.5x = 0.9589


0.1Ar0.9589O2.9589N+0.0823NO 22 ++

3 N2 1 O2

0.1 Ar 2400 K 10 atm


16-32

16-39 The mole fraction of sodium that ionizes according to the reaction Na ⇔ Na+ + e- at 2000 K and 0.8 atm is to be determined.

Assumptions All components behave as ideal gases.


Stoichiometric: )1 and 1 ,1 (thus e+NaNa -+ eNaNa-+ ===⇔ ννν

Actual: { 4434421products

+

react.

eNaNaNa −++⎯→⎯ yyx

Na balance: 1 1= + = −x y or y x

Total number of moles: N x y xtotal = + = −2 2

The equilibrium constant relation becomes,

111

total

2)(

totalNa

NaNa-e+Na

Na

-e-

Na −+−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

NP

xy

NP

N

NNK e

p

ννν

ν

νν

Substituting,

⎟⎠⎞

⎜⎝⎛

−−

=xx

x2

8.0)1(668.02

Solving for x,

x = 0.325

Thus the fraction of Na which dissociates into Na+ and e- is

1 - 0.325 = 0.675 or 67.5%

Na ⇔ Na+ + e- 2000 K 0.8 atm


16-33

16-40 Liquid propane enters a combustion chamber. The equilibrium composition of product gases and the rate of heat transfer from the combustion chamber are to be determined.

Assumptions 1 The equilibrium composition consists of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases.

Analysis (a) Considering 1 kmol of C3H8, the stoichiometric combustion equation can be written as

2th2222th83 N3.76+OH4CO3)N3.76(O)(HC aa +⎯→⎯++l

where ath is the stoichiometric coefficient and is determined from the O2 balance,

2.5 3 2 1.5 5th th tha a a= + + ⎯→⎯ =

Then the actual combustion equation with 150% excess air and some CO in the products can be written as

C H O N CO + (9 0.5 )O H O + 47N3 8 2 2 2 2 2 2( ) . ( . ) ( )COl + + ⎯→⎯ + − − +12 5 376 3 4x x x

After combustion, there will be no C3 H8 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as

) and ,1 ,1 (thus O+COCO 21

OCOCO221

2 22===⇔ ννν

and

)(

totalCO

OCO2CO2OCO

2CO

2

2O

2

CO ννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

where

N x x x xtotal = + − + − + + = −( ) ( . ) .3 9 0 5 4 47 63 0 5

From Table A-28, 81073.1 Thus K. 1200at 871.17ln −×=−= pp KK . Substituting,

15.15.0

8

5.0632)5.09)(3(1073.1

−− ⎟

⎠⎞

⎜⎝⎛

−−−

=×xx

xx

Solving for x,

0.39999999.2 ≅=x

Therefore, the amount CO in the product gases is negligible, and it can be disregarded with no loss in accuracy. Then the combustion equation and the equilibrium composition can be expressed as

C H O N CO O H O + 47N3 8 2 2 2 2 2 2( ) . ( . ) .l + + ⎯ →⎯ + +12 5 376 3 7 5 4

and

3CO 7.5O 4H O +47N2 2 2 2+ +

(b) The heat transfer for this combustion process is determined from the steady-flow energy balance E E Ein out system− = Δ on the combustion chamber with W = 0,

( ) ( )∑ ∑ −+−−+=−RfRPfP hhhNhhhNQ oooo

out

1200 K

C3H8

25°C

Air

12°C

Combustion chamber

2 atm

CO CO2 H2O O2 N2


16-34

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, (The hfo of liquid propane is obtained by adding the hfg at 25°C to hf

o of gaseous propane).

Substance

ofh

kJ/kmol

h285 K

kJ/kmol

h298 K

kJ/kmol

h1200 K

kJ/kmol

C3H8 (l) -118,910 --- --- ---

O2 0 8696.5 8682 38,447

N2 0 8286.5 8669 36,777

H2O (g) -241,820 --- 9904 44,380

CO2 -393,520 --- 9364 53,848

Substituting,

− = − + − + − + −+ + − + + −− − + − − + −− + −

= −

Q

h h

out

3 8 kJ / kmol of C H

3 393 520 53 848 9364 4 241 820 44 380 99047 5 0 38 447 8682 47 0 36 777 86691 118 910 12 5 0 8296 5 868247 0 8186 5 8669

185 764

298 298

( , , ) ( , , ). ( , ) ( , )( , ) . ( . )

( . ),

or

Qout 3 8 kJ / kmol of C H=185 764,

The mass flow rate of C3H8 can be expressed in terms of the mole numbers as

& & . .N mM

= = =1244

0 02727 kg / min kg / kmol

kmol / min

Thus the rate of heat transfer is

kJ/min 5066==×= kJ/kmol) 185,746kmol/min)( 02727.0(outout QNQ &&

The equilibrium constant for the reaction NOON 221

221 ⇔+ is ln Kp = -7.569, which is very small. This

indicates that the amount of NO formed during this process will be very small, and can be disregarded.


16-35

16-41 EES Problem 16-40 is reconsidered. It is to be investigated if it is realistic to disregard the presence of NO in the product gases.


"To solve this problem, the Gibbs function of the product gases is minimized. Click on the Min/Max icon." For this problem at 1200 K the moles of CO are 0.000 and moles of NO are 0.000, thus we can disregard both the CO and NO. However, try some product temperatures above 1286 K and observe the sign change on the Q_out and the amout of CO and NO present as the product temperature increases." "The reaction of C3H8(liq) with excess air can be written: C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO The coefficients A_th and EX are the theoretical oxygen and the percent excess air on a decimal basis. Coefficients a, b, c, d, e, and f are found by minimiming the Gibbs Free Energy at a total pressure of the product gases P_Prod and the product temperature T_Prod. The equilibrium solution can be found by applying the Law of Mass Action or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are six compounds present in the products subject to four specie balances, so there are two degrees of freedom. Minimize the Gibbs function of the product gases with respect to two molar quantities such as coefficients b and f. The equilibrium mole numbers a, b, c, d, e, and f will be determined and displayed in the Solution window." PercentEx = 150 [%] Ex = PercentEx/100 "EX = % Excess air/100" P_prod =2*P_atm T_Prod=1200 [K] m_dot_fuel = 0.5 [kg/s] Fuel$='C3H8' T_air = 12+273 "[K]" T_fuel = 25+273 "[K]" P_atm = 101.325 [kPa] R_u=8.314 [kJ/kmol-K] "Theoretical combustion of C3H8 with oxygen: C3H8 + A_th O2 = 3 C02 + 4 H2O " 2*A_th = 3*2 + 4*1 "Balance the reaction for 1 kmol of C3H8" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" b_max = 3 f_max = (1+Ex)*A_th*3.76*2 e_guess=Ex*A_th 1*3 = a*1+b*1 "Carbon balance" 1*8=c*2 "Hydrogen balance" (1+Ex)*A_th*2=a*2+b*1+c*1+e*2+f*1 "Oxygen balance" (1+Ex)*A_th*3.76*2=d*2+f*1 "Nitrogen balance"


16-36

"Total moles and mole fractions" N_Total=a+b+c+d+e+f y_CO2=a/N_Total; y_CO=b/N_Total; y_H2O=c/N_Total; y_N2=d/N_Total; y_O2=e/N_Total; y_NO=f/N_Total "The following equations provide the specific Gibbs function for each component as a function of its molar amount" g_CO2=Enthalpy(CO2,T=T_Prod)-T_Prod*Entropy(CO2,T=T_Prod,P=P_Prod*y_CO2) g_CO=Enthalpy(CO,T=T_Prod)-T_Prod*Entropy(CO,T=T_Prod,P=P_Prod*y_CO) g_H2O=Enthalpy(H2O,T=T_Prod)-T_Prod*Entropy(H2O,T=T_Prod,P=P_Prod*y_H2O) g_N2=Enthalpy(N2,T=T_Prod)-T_Prod*Entropy(N2,T=T_Prod,P=P_Prod*y_N2) g_O2=Enthalpy(O2,T=T_Prod)-T_Prod*Entropy(O2,T=T_Prod,P=P_Prod*y_O2) g_NO=Enthalpy(NO,T=T_Prod)-T_Prod*Entropy(NO,T=T_Prod,P=P_Prod*y_NO) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance" Gibbs=a*g_CO2+b*g_CO+c*g_H2O+d*g_N2+e*g_O2+f*g_NO "For the energy balance, we adjust the value of the enthalpy of gaseous propane given by EES:" h_fg_fuel = 15060"[kJ/kmol]" "Table A.27" h_fuel = enthalpy(Fuel$,T=T_fuel)-h_fg_fuel "Energy balance for the combustion process:" "C3H8(l) + (1+Ex)A_th (O2+3.76N2) = a C02 + b CO + c H2O + d N2 + e O2 + f NO" HR =Q_out+HP HR=h_fuel+ (1+Ex)*A_th*(enthalpy(O2,T=T_air)+3.76*enthalpy(N2,T=T_air)) HP=a*enthalpy(CO2,T=T_prod)+b*enthalpy(CO,T=T_prod)+c*enthalpy(H2O,T=T_prod)+d*enthalpy(N2,T=T_prod)+e*enthalpy(O2,T=T_prod)+f*enthalpy(NO,T=T_prod) "The heat transfer rate is:" Q_dot_out=Q_out/molarmass(Fuel$)*m_dot_fuel "[kW]" SOLUTION a=3.000 [kmol] A_th=5 b=0.000 [kmol] b_max=3 c=4.000 [kmol] d=47.000 [kmol] e=7.500 [kmol] Ex=1.5 e_guess=7.5 f=0.000 [kmol] Fuel$='C3H8' f_max=94 Gibbs=-17994897 [kJ] g_CO=-703496 [kJ/kmol]

g_CO2=-707231 [kJ/kmol] g_H2O=-515974 [kJ/kmol] g_N2=-248486 [kJ/kmol] g_NO=-342270 [kJ/kmol] g_O2=-284065 [kJ/kmol] HP=-330516.747 [kJ/kmol] HR=-141784.529 [kJ/kmol] h_fg_fuel=15060 [kJ/kmol] h_fuel=-118918 [kJ/kmol] m_dot_fuel=0.5 [kg/s] N_Total=61.5 [kmol/kmol_fuel] PercentEx=150 [%] P_atm=101.3 [kPa] P_prod=202.7 [kPa]

Q_dot_out=2140 [kW] Q_out=188732 [kJ/kmol_fuel] R_u=8.314 [kJ/kmol-K] T_air=285 [K] T_fuel=298 [K] T_Prod=1200.00 [K] y_CO=1.626E-15 y_CO2=0.04878 y_H2O=0.06504 y_N2=0.7642 y_NO=7.857E-08 y_O2=0.122


16-37

16-42 Oxygen is heated during a steady-flow process. The rate of heat supply needed during this process is to be determined for two cases. Assumptions 1 The equilibrium composition consists of O2 and O. 2 All components behave as ideal gases. Analysis (a) Assuming some O2 dissociates into O, the dissociation equation can be written as

O O ( )O2 2 2 1⎯→⎯ + −x x

The equilibrium equation among O2 and O can be expressed as )2 and 1 (thus 2OO OO2 2

==⇔ νν

Assuming ideal gas behavior for all components, the equilibrium constant relation can be expressed as

2OO

2O

2

O

totalO

Oνν

ν

ν −

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK p

where N x x xtotal = + − = −2 1 2( )

From Table A-28, .01282.0 Thus K. 3000at 357.4ln =−= pp KK Substituting,

122

21)22(01282.0

−

⎟⎠⎞

⎜⎝⎛

−−

=xx

x

Solving for x gives x = 0.943 Then the dissociation equation becomes

O O O2 2⎯→⎯ +0 943 0114. .

The heat transfer for this combustion process is determined from the steady-flow energy balance E E Ein out system− = Δ on the combustion chamber with W = 0,

( ) ( )∑ ∑ −+−−+=RfRPfP hhhNhhhNQ oooo

in

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

Substance

hfo

kJ/kmol

h298 K

kJ/kmol

h3000 K

kJ/kmol O 249,190 6852 63,425 O2 0 8682 106,780

Substituting, Qin 2 kJ / kmol O= + − + + − − =0 943 0 106 780 8682 0114 249 190 63 425 6852 0 127 363. ( , ) . ( , , ) ,

The mass flow rate of O2 can be expressed in terms of the mole numbers as

kmol/min 01563.0kg/kmol 32kg/min 5.0

===MmN&&

Thus the rate of heat transfer is

kJ/min 1990==×= kJ/kmol) 127,363kmol/min)( 01563.0(inin QNQ &&

(b) If no O2 dissociates into O, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

kJ/min 1533==−=−= kJ/kmol 8682)-106,780kmol/min)( 01563.0()()( 1212in hhNhhmQ &&&

O2, O

3000 K

Q&

O2

298 K


16-38

16-43 The equilibrium constant, Kp is to be estimated at 2500 K for the reaction CO + H2O = CO2 + H2. Analysis (a) The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

TRTGKeK upTRTG

pu /)(*lnor /)(* Δ−== Δ−

where

)()()()()(* H2OH2OCOCOH2H2CO2CO2 TgTgTgTgTG ∗∗∗∗ −−+=Δ νννν

At 2500 K,

[ ] [ ][ ] [ ]

kJ/kmol 525,37)18.276)(2500()891,142(1)65.266)(2500()510,35(1

)10.196)(2500()452,70(1)60.322)(2500()641,271(1)()()()(

)()()()()(*

H2OH2OCOCOH2H2CO2CO2

H2OH2OCOCOH2H2CO2CO2

=−−−−−−

−+−−=−−−−−+−=

−−+=Δ ∗∗∗∗

sThsThsThsTh

TgTgTgTgTG

νννν

νννν

The enthalpies at 2500 K and entropies at 2500 K and 101.3 kPa (1 atm) are obtained from EES. Substituting,

0.1644=⎯→⎯−=⋅

−= pp KK 8054.1K) K)(2500kJ/kmol (8.314

kJ/kmol 525,37ln

The equilibrium constant may be estimated using the integrated van't Hoff equation:

0.1612=⎯→⎯⎟⎠⎞

⎜⎝⎛ −

−=⎟

⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛

est,est,

1

est,

K 25001

K 20001

kJ/kmol.K 314.8kJ/kmol 176,26

2209.0ln

11ln

pp

Ru

R

p

p

KK

TTRh

KK


16-39

16-44 A constant volume tank contains a mixture of H2 and O2. The contents are ignited. The final temperature and pressure in the tank are to be determined.

Analysis The reaction equation with products in equilibrium is

22222 O OH H OH cba ++⎯→⎯+

The coefficients are determined from the mass balances

Hydrogen balance: ba 222 +=

Oxygen balance: cb 22 +=


222 O5.0HOH +⎯→←



p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(* prodH2OH2OprodO2O2prodH2H2 TgTgTgTG ∗∗∗ −+=Δ ννν

and the Gibbs functions are given by

H2OprodprodH2O

O2prodprodO2

H2prodprodH2

)()(

)()(

)()(

sThTg

sThTg

sThTg

−=

−=

−=

∗

∗

∗


5.0

25.015.01

tot1

5.01 3.101/⎟⎟⎠

⎞⎜⎜⎝

⎛++

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

−+

cbaP

bac

NP

bcaK p

An energy balance on the tank under adiabatic conditions gives

PR UU =

where

kJ/kmol 4958K) (298.15kJ/kmol.K) 314.8(0K) (298.15kJ/kmol.K) 314.8(0

)(1)(1 reacCO2@25reacCH2@25

−=−+−=

−+−= °° TRhTRhU uuR

)()()( prod O2@prod H2O@prod H2@ prodprodprodTRhcTRhbTRhaU uTuTuTP −+−+−=

The relation for the final pressure is

kPa) 3.101(K 15.2982

prod1

reac

prod

1

tot2 ⎟

⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ++

==TcbaP

TT

NN

P

Solving all the equations simultaneously using EES, we obtain the final temperature and pressure in the tank to be

kPa 1043K 3857

=

=

2

prod

P

T


16-40

Simultaneous Reactions

16-45C It can be expresses as “(dG)T,P = 0 for each reaction.” Or as “the Kp relation for each reaction must be satisfied.”

16-46C The number of Kp relations needed to determine the equilibrium composition of a reacting mixture is equal to the difference between the number of species present in the equilibrium mixture and the number of elements.


16-41

16-47 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined.

Assumptions 1 The equilibrium composition consists of H2O, OH, O2, and H2. 2 The constituents of the mixture are ideal gases.

Analysis The reaction equation during this process can be expressed as

H O H O H O + OH2 2 2 2⎯→⎯ + +x y z w

Mass balances for hydrogen and oxygen yield

H balance: 2 2 2= + +x y w (1)

O balance: 1 2= + +x z w (2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the Kp relations) to determine the equilibrium composition of the mixture. They are

221

22 OHOH +⇔ (reaction 1)

OHHOH 221

2 +⇔ (reaction 2)

The equilibrium constant for these two reactions at 3400 K are determined from Table A-28 to be

ln . .

ln . .

K K

K KP P

P P

1 1

2 2

1891 015092

1576 0 20680

= − ⎯→⎯ =

= − ⎯→⎯ =

The Kp relations for these two simultaneous reactions are

)(

totalOH

OHH2

)(

totalOH

OH1

O2HOH2H

O2H

2

OH2H

2O2H2O2H

O2H

2

2O

2

2H

2 and ννν

ν

ννννν

ν

νν −+−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK

NP

N

NNK PP

where N N N N N x y z wtotal H O H O OH2 2 2= + + + = + + +

Substituting,

2/12/1 1))((15092.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

zy (3)

2/12/1 1))((20680.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

yw (4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields

x = 0.574 y = 0.308 z = 0.095 w = 0.236

Therefore, the equilibrium composition becomes

0.574H O 0.308H 0.095O 0.236OH2 2 2+ + +

2H,2O

OHO,2HO2H ⇒

3400 K 1 atm


16-42

16-48 Two chemical reactions are occurring in a mixture. The equilibrium composition at a specified temperature is to be determined.

Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases.


2CO + O CO CO O + O2 2 2 2⎯→⎯ + +x y z w

Mass balances for carbon and oxygen yield

C balance: 2 = +x y (1)

O balance: 6 2 2= + + +x y z w (2)

The mass balances provide us with only two equations with four unknowns, and thus we need to have two more equations (to be obtained from the KP relations) to determine the equilibrium composition of the mixture. They are

221

2 OCOCO +⇔ (reaction 1)

O 2O2 ⇔ (reaction 2)


ln . .

ln . .

K K

K KP P

P P

1 1

2 2

0 429 0 65116

3072 0 04633

= − ⎯→⎯ =

= − ⎯→⎯ =

The KP relations for these two simultaneous reactions are

2OO

2O

2

O

2CO2OCO

2CO

2

2O

2

CO

totalO

O2

)(

totalCO

OCO1

νν

ν

ν

ννν

ν

νν

−

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK

NP

N

NNK

P

P

where

N N N N N x y z wtotal CO O CO O2 2= + + + = + + +

Substituting,

2/12/1 2))((65116.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

zy (3)

122 204633.0−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxz

w (4)


x = 1.127 y = 0.873 z = 1.273 w = 0.326

Thus the equilibrium composition is

0.326O1.273O0.873CO1.127CO 22 +++

CO2, CO, O2, O 3200 K 2 atm


16-43

16-49 Two chemical reactions are occurring at high-temperature air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases.


O + 3.76 N N NO O + O2 2 2 2⎯→⎯ + +x y z w

Mass balances for nitrogen and oxygen yield

N balance: yx += 252.7 (1)

O balance: wzy ++= 22 (2)


12

12N O NO2 2+ ⇔ (reaction 1)



ln . .

ln . .

K K

K KP P

P P

1 1

2 2

2114 012075

4 357 0 01282

= − ⎯→⎯ =

= − ⎯→⎯ =


2OO

2O

2

O

2O2NNO

2O

2

2N

2

NO

totalO

O2

)(

totalON

NO1

νν

ν

ν

ννν

νν

ν

−

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK

NP

NN

NK

P

P

where wzyxNNNNN +++=+++= OONONtotal 22

Substituting,

5.05.01

5.05.0212075.0

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxzx

y (3)

122 201282.0−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxz

w (4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.656 y = 0.2086 z = 0.8162 w = 0.1591 Thus the equilibrium composition is

0.1591O0.8162O0.2086NO3.656N 22 +++

The equilibrium constant of the reaction N 2N2 ⇔ at 3000 K is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

O2, N2, O, NO

3000 K

Heat

Reaction chamber, 2 atm

AIR


16-44

16-50E [Also solved by EES on enclosed CD] Two chemical reactions are occurring in air. The equilibrium composition at a specified temperature is to be determined. Assumptions 1 The equilibrium composition consists of O2, N2, O, and NO. 2 The constituents of the mixture are ideal gases. Analysis The reaction equation during this process can be expressed as

O + 3.76 N N NO O O2 2 2 2⎯→⎯ + + +x y z w

Mass balances for nitrogen and oxygen yield

N balance: 752 2. = +x y (1)

O balance: 2 2= + +y z w (2)


NOON 221

221 ⇔+ (reaction 1)


The equilibrium constant for these two reactions at T = 5400 R = 3000 K are determined from Table A-28 to be

ln . .

ln . .

K K

K KP P

P P

1 1

2 2

2114 012075

4 357 0 01282

= − ⎯ →⎯ =

= − ⎯ →⎯ =


2OO

2O

2

O

2O2NNO

2O

2

2N

2

NO

totalO

O2

)(

totalON

NO1

νν

ν

ν

ννν

νν

ν

−

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK

NP

NN

NK

P

P

where N N N N N x y z wtotal N NO O O2 2= + + + = + + +

Substituting,

5.05.01

5.05.0112075.0

−−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxzx

y (3)

122 101282.0−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxz

w (4)

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y, z, and w yields x = 3.658 y = 0.2048 z = 0.7868 w = 0.2216 Thus the equilibrium composition is

0.2216O0.7868O0.2048NO3.658N 22 +++

The equilibrium constant of the reaction N 2N2 ⇔ at 5400 R is lnKP = -22.359, which is much smaller than the KP values of the reactions considered. Therefore, it is reasonable to assume that no N will be present in the equilibrium mixture.

O2, N2, O, NO

5400 R

Heat

Reaction chamber, 1 atm

AIR


16-45

14-51E EES Problem 16-50E is reconsidered. Using EES (or other) software, the equilibrium solution is to be obtained by minimizing the Gibbs function by using the optimization capabilities built into EES. This solution technique is to be compared with that used in the previous problem. Analysis The problem is solved using EES, and the solution is given below.

"This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. 0.21 O2+0.79 N2 = a O2+b O + c N2 + d NO Two of the four coefficients, a, b, c, and d, are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 5400 R. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, select MinMax from the Calculate menu. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficients b and d. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "Data from Data Input Window" {T=5400 "R" P=1 "atm" } AO2=0.21; BN2=0.79 "Composition of air" AO2*2=a*2+b+d "Oxygen balance" BN2*2=c*2+d "Nitrogen balance" "The total moles at equilibrium are" N_tot=a+b+c+d y_O2=a/N_tot; y_O=b/N_tot; y_N2=c/N_tot; y_NO=d/N_tot "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T)-T*Entropy(O2,T=T,P=P*y_O2) g_N2=Enthalpy(N2,T=T)-T*Entropy(N2,T=T,P=P*y_N2) g_NO=Enthalpy(NO,T=T)-T*Entropy(NO,T=T,P=P*y_NO) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ so we must convert h and s to English units." T_K=T*Convert(R,K) "Convert R to K" Call JANAF('O',T_K:Cp`,h`,S`) "Units from JANAF are SI" S_O=S`*Convert(kJ/kgmole-K, Btu/lbmole-R) h_O=h`*Convert(kJ/kgmole, Btu/lbmole) "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_O=h_O-T*(S_O-R_u*ln(Y_O)) R_u=1.9858 "The universal gas constant in Btu/mole-R " "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=a*g_O2+b*g_O+c*g_N2+d*g_NO


16-46

d

[lbmol] b

[lbmol] Gibbs

[Btu/lbmol]yO2 yO yNO yN2 T

[R] 0.002698 0.00001424 -162121 0.2086 0.0000 0.0027 0.7886 3000 0.004616 0.00006354 -178354 0.2077 0.0001 0.0046 0.7877 3267 0.007239 0.0002268 -194782 0.2062 0.0002 0.0072 0.7863 3533 0.01063 0.000677 -211395 0.2043 0.0007 0.0106 0.7844 3800 0.01481 0.001748 -228188 0.2015 0.0017 0.0148 0.7819 4067 0.01972 0.004009 -245157 0.1977 0.0040 0.0197 0.7786 4333 0.02527 0.008321 -262306 0.1924 0.0083 0.0252 0.7741 4600 0.03132 0.01596 -279641 0.1849 0.0158 0.0311 0.7682 4867 0.03751 0.02807 -297179 0.1748 0.0277 0.0370 0.7606 5133 0.04361 0.04641 -314941 0.1613 0.0454 0.0426 0.7508 5400

3000 3500 4000 4500 5000 55000.000

0.010

0.020

0.030

0.040

0.050

T [R]

Mol

e fr

actio

n of

NO

and

O

NO

O

Discussion The equilibrium composition in the above table are based on the reaction in which the reactants are 0.21 kmol O2 and 0.79 kmol N2. If you multiply the equilibrium composition mole numbers above with 4.76, you will obtain equilibrium composition for the reaction in which the reactants are 1 kmol O2 and 3.76 kmol N2.This is the case in problem 16-43E.


16-47

16-52 Water vapor is heated during a steady-flow process. The rate of heat supply for a specified exit temperature is to be determined for two cases.


Analysis (a) Assuming some H2O dissociates into H2, O2, and O, the dissociation equation can be written as

H O H O H O + OH2 2 2 2⎯→⎯ + +x y z w


H balance: 2 2 2= + +x y w (1)

O balance: 1 2= + +x z w (2)


221

22 OHOH +⇔ (reaction 1)

OHHOH 221

2 +⇔ (reaction 2)


ln . .

ln . .

K K

K KP P

P P

1 1

2 2

3086 0 04568

2 937 0 05302

= − ⎯ →⎯ =

= − ⎯ →⎯ =

The KP relations for these three simultaneous reactions are

)(

totalOH

OHH2

)(

totalOH

OH1

O2H2O2H

O2H

2

OH2H

2

O2H2O2H

O2H

2

2O

2

2H

2

ννν

ν

νν

ννν

ν

νν

−+

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK

NP

N

NNK

P

P

where

wzyxNNNNN +++=+++= OHOHOHtotal 222

Substituting,

2/12/1 1))((04568.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

zy (3)

2/12/1 1))((05302.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

yw (4)


x = 0.784 y = 0.162 z = 0.054 w = 0.108

Thus the balanced equation for the dissociation reaction is

H O 0.784H O 0.162H 0.054O 0.108OH2 2 2 2⎯ →⎯ + + +

H2O, H2, O2, OH

3000 K

Q

H2O

298 K


16-48

The heat transfer for this dissociation process is determined from the steady-flow energy balance E E Ein out system− = Δ with W = 0,

( ) ( )∑ ∑ −+−−+=RfRPfP hhhNhhhNQ oooo

in

Assuming the O2 and O to be ideal gases, we have h = h(T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h3000 K

kJ/kmol

H2O -241,820 9904 136,264

H2 0 8468 97,211

O2 0 8682 106,780

OH 39,460 9188 98,763

Substituting,

Qin

2

kJ / kmol H O

= − + −+ + −+ + −+ + − − −

=

0 784 241 820 136 264 99040162 0 97 211 84680 054 0 106 780 86820108 39 460 98 763 9188 241 820

184 909

. ( , , ). ( , ). ( , ). ( , , ) ( , ),

The mass flow rate of H2O can be expressed in terms of the mole numbers as

& & . .N mM

= = =0 218

0 01111 kg / min kg / kmol

kmol / min

Thus,

kJ/min 2055==×= kJ/kmol) 184,909kmol/min)( 01111.0(inin QNQ &&

(b) If no dissociates takes place, then the process involves no chemical reactions and the heat transfer can be determined from the steady-flow energy balance for nonreacting systems to be

& & ( ) & ( )( .

Q m h h N h hin

kmol / min)(136,264 ) kJ / kmol= − = −= −=

2 1 2 1

0 01111 99041404 kJ / min


16-49

16-53 EES Problem 16-52 is reconsidered. The effect of the final temperature on the rate of heat supplied for the two cases is to be studied.


"This example illustrates how EES can be used to solve multi-reaction chemical equilibria problems by directly minimizing the Gibbs function. H2O = x H2O+y H2+z O2 + w OH Two of the four coefficients, x, y, z, and w are found by minimiming the Gibbs function at a total pressure of 1 atm and a temperature of 3000 K. The other two are found from mass balances. The equilibrium solution can be found by applying the Law of Mass Action (Eq. 15-15) to two simultaneous equilibrium reactions or by minimizing the Gibbs function. In this problem, the Gibbs function is directly minimized using the optimization capabilities built into EES. To run this program, click on the Min/Max icon. There are four compounds present in the products subject to two elemental balances, so there are two degrees of freedom. Minimize Gibbs with respect to two molar quantities such as coefficient z and w. The equilibrium mole numbers of each specie will be determined and displayed in the Solution window. Minimizing the Gibbs function to find the equilibrium composition requires good initial guesses." "T_Prod=3000 [K]" P=101.325 [kPa] m_dot_H2O = 0.2 [kg/min] T_reac = 298 [K] T = T_prod P_atm=101.325 [kPa] "H2O = x H2O+y H2+z O2 + w OH" AH2O=1 "Solution for 1 mole of water" AH2O=x+z*2+w "Oxygen balance" AH2O*2=x*2+y*2+w "Hydrogen balance" "The total moles at equilibrium are" N_tot=x+y+z+w y_H2O=x/N_tot; y_H2=y/N_tot; y_O2=z/N_tot; y_OH=w/N_tot "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kgmole, K, and kJ." Call JANAF('OH',T_prod:Cp`,h`,S`) "Units from JANAF are SI" S_OH=S` h_OH=h` "The entropy from JANAF is for one atmosphere so it must be corrected for partial pressure." g_OH=h_OH-T_prod*(S_OH-R_u*ln(y_OH*P/P_atm)) R_u=8.314 "The universal gas constant in kJ/kmol-K " "The following equations provide the specific Gibbs function for three of the components." g_O2=Enthalpy(O2,T=T_prod)-T_prod*Entropy(O2,T=T_prod,P=P*y_O2) g_H2=Enthalpy(H2,T=T_prod)-T_prod*Entropy(H2,T=T_prod,P=P*y_H2)


16-50

g_H2O=Enthalpy(H2O,T=T_prod)-T_prod*Entropy(H2O,T=T_prod,P=P*y_H2O) "The extensive Gibbs function is the sum of the products of the specific Gibbs function and the molar amount of each substance." Gibbs=x*g_H2O+y*g_H2+z*g_O2+w*g_OH "H2O = x H2O+y H2+z O2 + w OH" 1*Enthalpy(H2O,T=T_reac)+Q_in=x*Enthalpy(H2O,T=T_prod)+y*Enthalpy(H2,T=T_prod)+z*Enthalpy(O2,T=T_prod)+w*h_OH N_dot_H2O = m_dot_H2O/molarmass(H2O) Q_dot_in_Dissoc = N_dot_H2O*Q_in Q_dot_in_NoDissoc = N_dot_H2O*(Enthalpy(H2O,T=T_prod) - Enthalpy(H2O,T=T_reac))

Tprod [K] Qin,Dissoc [kJ/min]

Qin,NoDissoc [kJ/min]

2500 1266 1098 2600 1326 1158 2700 1529 1219 2800 1687 1280 2900 1862 1341 3000 2053 1403 3100 2260 1465 3200 2480 1528 3300 2710 1590 3400 2944 1653 3500 3178 1716

2500 2700 2900 3100 3300 35001000

1500

2000

2500

3000

TProd [K]

Qin

[kJ/

min

]

Qin,Dissoc

Qin,NoDissoc


16-51

16-54 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted.

Analysis The complete combustion reaction in this case can be written as

[ ] 22th2222th52 N O ))((OH 3CO 23.76NO)1((gas) OHHC faExaEx +++⎯→⎯+++

where ath is the stoichiometric coefficient for air. The oxygen balance gives

2))((13222)1(1 thth ×+×+×=×++ aExaEx

The reaction equation with products in equilibrium is

[ ] NO N O OH CO CO 3.76NO)1((gas) OHHC 222222th52 gfedbaaEx +++++⎯→⎯+++


Carbon balance: ba +=2

Hydrogen balance: 326 =⎯→⎯= dd

Oxygen balance: gedbaaEx +×+++×=×++ 222)1(1 th

Nitrogen balance: gfaEx +×=××+ 2276.3)1( th

Solving the above equations, we find the coefficients to be

Ex = 0.4, ath = 3, a = 1.995, b = 0.004712, d = 3, e = 1.17, f = 15.76, g = 0.06428

Then, we write the balanced reaction equation as

[ ]NO 06428.0N 76.15O 17.1OH 3CO 004712.0CO 995.1

3.76NO2.4(gas) OHHC

2222

2252

+++++⎯→⎯

++


99.2176.1517.13004712.0995.1tot =++++=N

The first assumed equilibrium reaction is

22 O5.0COCO +⎯→←


⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ−=

prod

prod11

)(*exp

TRTG

Ku

p

Where )()()()(* prodCO2CO2prodO2O2prodCOCOprod1 TgTgTgTG ∗∗∗ −+=Δ ννν


CO2prodprodCO2

O2prodprodO2

COprodprodCO

)()(

)()(

)()(

sThTg

sThTg

sThTg

−=

−=

−=

∗

∗

∗



16-52

0005447.099.21

1995.1

)17.1)(004712.0( 5.05.015.01

tot

5.0

1 =⎟⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−+

NP

abeK p

The second assumed equilibrium reaction is

NOO5.0N5.0 22 ⎯→←+

Also, for this reaction, we have

N2prodprodN2

NOprodprodNO

)()(

)()(

sThTg

sThTg

−=

−=∗

∗

O2prodprodO2 )()( sThTg −=∗

)()()()(* prodO2O2prodN2N2prodNONOprod2 TgTgTgTG ∗∗∗ −−=Δ ννν

⎟⎟⎠

⎞⎜⎜⎝

⎛ Δ−=

prod

prod22

)(*exp

TRTG

Ku

p

01497.0)76.15()17.1(

06428.099.21

15.05.0

0

5.05.

5.05.01

tot2 =⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−

feg

NPK

op

A steady flow energy balance gives

PR HH =

where

kJ/kmol 310,235.79(0)15(0)2.4kJ/kmol) 310,235(

79.152.4 CN2@25CO2@25Cfuel@25

−=++−=

++= °°°hhhH o

fR

prodprod

prodprodprodprod

NO@N2@

O2@H2O@CO@CO2@

06428.076.15

17.13004712.0995.1

TT

TTTTP

hh

hhhhH

++

+++=

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature

K 1901=prodT

The copy of entire EES solution including parametric studies is given next: "The reactant temperature is:" T_reac= 25+273 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O + Ex*A_th O2 + f N2 " "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2 + g NO"


16-53

"Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 +g "Nitrogen Balance:" (1+Ex)*A_th*3.76 *2= f*2 + g N_tot =a +b + d + e + f +g "Total kilomoles of products at equilibrium" "The first assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P_1 = exp(-DELTAG_1 /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot) *b *sqrt(e) =K_P_1*a "The econd assumed equilibrium reaction is 0.5N2+0.5O2=NO" g_NO=Enthalpy(NO,T=T_prod )-T_prod *Entropy(NO,T=T_prod ,P=101.3) g_N2=Enthalpy(N2,T=T_prod )-T_prod *Entropy(N2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG_2 =1*g_NO-0.5*g_O2-0.5*g_N2 "The equilibrium constant is given by Eq. 15-14." K_P_2 = exp(-DELTAG_2 /(R_u*T_prod )) "The equilibrium constant is also given by Eq. 15-15." "K_ P_2 = (P/N_tot)^(1-0.5-0.5)*(g^1)/(e^0.5*f^0.5)" g=K_P_2 *sqrt(e*f) "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod)+e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod)+g*ENTHALPY(NO,T=T_prod) "[kJ/kmol]"


16-54

ath a b d e f g PercentEx [%]

Tprod [K]

3 1.922 0.07779 3 0.3081 12.38 0.0616 10 2184 3 1.971 0.0293 3 0.5798 13.5 0.06965 20 2085 3 1.988 0.01151 3 0.8713 14.63 0.06899 30 1989 3 1.995 0.004708 3 1.17 15.76 0.06426 40 1901 3 1.998 0.001993 3 1.472 16.89 0.05791 50 1820 3 1.999 0.0008688 3 1.775 18.02 0.05118 60 1747 3 2 0.0003884 3 2.078 19.15 0.04467 70 1682 3 2 0.0001774 3 2.381 20.28 0.03867 80 1621 3 2 0.00008262 3 2.683 21.42 0.0333 90 1566 3 2 0.00003914 3 2.986 22.55 0.02856 100 1516

10 20 30 40 50 60 70 80 90 1001500

1600

1700

1800

1900

2000

2100

2200

PercentEx

T pro

d (K

)


16-55

Variations of Kp with Temperature

16-55C It enables us to determine the enthalpy of reaction hR from a knowledge of equilibrium constant KP.

16-56C At 2000 K since combustion processes are exothermic, and exothermic reactions are more complete at lower temperatures.


16-56

16-57 The hR at a specified temperature is to be determined using the enthalpy and KP data.

Assumptions Both the reactants and products are ideal gases.

Analysis (a) The complete combustion equation of CO can be expressed as

2221 COO+CO ⇔

The hR of the combustion process of CO at 2200 K is the amount of energy released as one kmol of CO is burned in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from

( ) ( )∑ ∑ −+−−+=RfRPfPR hhhNhhhNh oooo

Assuming the CO, O2 and CO2 to be ideal gases, we have h = h(T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2200 K

kJ/kmol

CO2 -393,520 9364 112,939

CO -110,530 8669 72,688

O2 0 8682 75,484

Substituting,

hR = − + −− − + −− + −

= −

1 393 520 112 939 93641 110 530 72 688 86690 5 0 75 484 8682

( , , )( , , ). ( , )

276,835 kJ / kmol

(b) The hR value at 2200 K can be estimated by using KP values at 2000 K and 2400 K (the closest two temperatures to 2200 K for which KP data are available) from Table A-28,

⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

kJ/kmol 276,856−≅

⎟⎠⎞

⎜⎝⎛ −

⋅≅−

R

R

h

hK 2400

1K 2000

1KkJ/kmol 314.8

635.6860.3


16-57

16-58E The hR at a specified temperature is to be determined using the enthalpy and KP data.


Analysis (a) The complete combustion equation of CO can be expressed as

2221 COO+CO ⇔

The hR of the combustion process of CO at 3960 R is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 3960 R, and can be determined from


Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables,

Substance hfo

Btu/lbmol

h537 R

Btu/lbmol

h3960 R

Btu/lbmol

CO2 -169,300 4027.5 48,647

CO -47,540 3725.1 31,256.5

O2 0 3725.1 32,440.5

Substituting,

hR = − + −− − + −− + −

= −

1 169 300 48 647 4027 51 47 540 31 256 5 372510 5 0 32 440 5 37251

( , , . )( , , . . ). ( , . . )

119,030 Btu / lbmol

(b) The hR value at 3960 R can be estimated by using KP values at 3600 R and 4320 R (the closest two temperatures to 3960 R for which KP data are available) from Table A-28,

⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

Btu/lbmol 119,041−≅

⎟⎠⎞

⎜⎝⎛ −

⋅≅−

R

R

h

hR 4320

1R 3600

1RBtu/lbmol 986.1

635.6860.3


16-58

16-59 The KP value of the combustion process H2 + 1/2O2 ⇔ H2O is to be determined at a specified temperature using hR data and KP value .


Analysis The hR and KP data are related to each other by

⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

The hR of the specified reaction at 2400 K is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from


Assuming the H2O, H2 and O2 to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2400 K

kJ/kmol

H2O -241,820 9904 103,508

H2 0 8468 75,383

O2 0 8682 83,174

Substituting,

hR = − + −− + −− + −

= −

1 241 820 103 508 99041 0 75 383 84680 5 0 83 174 8682

252 377

( , , )( , ). ( , )

, kJ / kmol

The KP value at 2600 K can be estimated from the equation above by using this hR value and the KP value at 2200 K which is ln KP1 = 6.768,

⎟⎠⎞

⎜⎝⎛ −

⋅−

≅−K 2600

1K 2200

1KkJ/kmol 314.8

kJ/kmol 377,252768.6ln 2PK

ln . . )K KP P2 24 645 4 648= = (Table A - 28: ln

or

KP2 = 104.1


16-59

16-60 The hR value for the dissociation process CO2 ⇔ CO + 1/2O2 at a specified temperature is to be determined using enthalpy and Kp data.


Analysis (a) The dissociation equation of CO2 can be expressed as

221

2 O+COCO ⇔

The hR of the dissociation process of CO2 at 2200 K is the amount of energy absorbed or released as one kmol of CO2 dissociates in a steady-flow combustion chamber at a temperature of 2200 K, and can be determined from


Assuming the CO, O2 and CO2 to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2 00 K2

kJ/kmol

CO2 -393,520 9364 112,939

CO -110,530 8669 72,688

O2 0 8682 75,484

Substituting,

hR = − + −+ + −− − + −

=

1 110 530 72 688 86690 5 0 75 484 86821 393 520 112 939 9364

( , , ). ( , )( , , )

276,835 kJ / kmol


⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

kJ/kmol 276,856≅

⎟⎠⎞

⎜⎝⎛ −

⋅≅−−−

R

R

h

hK 2400

1K 2000

1KkJ/kmol 314.8

)635.6(860.3


16-60

16-61 The hR value for the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using enthalpy and KP data.


Analysis (a) The dissociation equation of O2 can be expressed as

O 2O2 ⇔

The hR of the dissociation process of O2 at 3100 K is the amount of energy absorbed or released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 3100 K, and can be determined from


Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2900 K

kJ/kmol

O 249,190 6852 65,520

O2 0 8682 110,784

Substituting,

kJ/kmol 513,614=

−+−−+= )8682784,1100(1)6852520,65190,249(2Rh


⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

kJ/kmol 512,808≅

⎟⎠⎞

⎜⎝⎛ −

⋅≅−−−

R

R

h

hK 3200

1K 3000

1KkJ/kmol 314.8

)357.4(072.3


16-61

16-62 The enthalpy of reaction for the equilibrium reaction CH4 + 2O2 = CO2 + 2H2O at 2500 K is to be estimated using enthalpy data and equilibrium constant, Kp data.

Analysis The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data using

TRTGKeK upTRTG

pu /)(*lnor /)(* Δ−== Δ−

where

)()()()()(* O2O2CH4CH4H2OH2OCO2CO2 TgTgTgTgTG ∗∗∗∗ −−+=Δ νννν

At T1 = 2500 - 10 = 2490 K:

kJ/kmol 929,794)582,611(2)973,717(1)577,830(2)10075.1(1

)()()()()(*6

1O2O21CH4CH41H2OH2O1CO2CO21

−=−−−−−+×−=

−−+=Δ ∗∗∗∗ TgTgTgTgTG νννν

At T2 = 2500 + 10 = 2510 K:

kJ/kmol 801,794)124,617(2)516,724(1)100,836(2)10081.1(1

)()()()()(*6

2O2O22CH4CH42H2OH2O2CO2CO22

−=−−−−−+×−=

−−+=Δ ∗∗∗∗ TgTgTgTgTG νννν

The Gibbs functions are obtained from enthalpy and entropy properties using EES. Substituting,

161 10747.4

K) K)(2490kJ/kmol (8.314kJ/kmol 929,794exp ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

−−=pK

162 10475.3

K) K)(2510kJ/kmol (8.314kJ/kmol 801,794exp ×=⎟⎟

⎠

⎞⎜⎜⎝

⎛⋅

−−=pK

The enthalpy of reaction is determined by using the integrated van't Hoff equation:

kJ/kmol 810,845−=⎯→⎯⎟⎠⎞

⎜⎝⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛

×

×

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛

RR

u

R

p

p

hh

TTRh

KK

K 25101

K 24901

kJ/kmol.K 314.810747.410475.3ln

11ln

16

16

211

2

The enthalpy of reaction can also be determined from an energy balance to be

RPR HHh −=

where

kJ/kmol 423,557)891,142(2)641,271(21

kJ/kmol 422,253)377,78(2668,9621

K 2500 @ H2OK 2500 @ CO2

K 2500 @ O2K 2500 @ CH4

−=−+−=+=

=+=+=

hhH

hhH

P

R

The enthalpies are obtained from EES. Substituting,

kJ/kmol 810,845−=−−=−= )422,253()423,557(RPR HHh

which is identical to the value obtained using Kp data.


16-62

Phase Equilibrium

16-63C No. Because the specific gibbs function of each phase will not be affected by this process; i.e., we will still have gf = gg.

16-64C Yes. Because the number of independent variables for a two-phase (PH=2), two-component (C=2) mixture is, from the phase rule,

IV = C - PH + 2 = 2 - 2 + 2 = 2

Therefore, two properties can be changed independently for this mixture. In other words, we can hold the temperature constant and vary the pressure and still be in the two-phase region. Notice that if we had a single component (C=1) two phase system, we would have IV=1, which means that fixing one independent property automatically fixes all the other properties.

11-65C Using solubility data of a solid in a specified liquid, the mass fraction w of the solid A in the liquid at the interface at a specified temperature can be determined from

liquidsolid

solidmfmm

mA +=

where msolid is the maximum amount of solid dissolved in the liquid of mass mliquid at the specified temperature.

11-66C The molar concentration Ci of the gas species i in the solid at the interface Ci, solid side (0) is proportional to the partial pressure of the species i in the gas Pi, gas side(0) on the gas side of the interface, and is determined from

)0(S)0( side gas i,side solid i, PC ×= (kmol/m3)

where S is the solubility of the gas in that solid at the specified temperature.

11-67C Using Henry’s constant data for a gas dissolved in a liquid, the mole fraction of the gas dissolved in the liquid at the interface at a specified temperature can be determined from Henry’s law expressed as

yP

Hi, liquid sidei, gas side( )

( )0

0=

where H is Henry’s constant and Pi,gas side(0) is the partial pressure of the gas i at the gas side of the interface. This relation is applicable for dilute solutions (gases that are weakly soluble in liquids).


16-63

16-68 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 100°C satisfies the criterion for phase equilibrium.

Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-4,

kJ/kg 62.68K)kJ/kg K)(7.3542 15.373(kJ/kg) 6.2675(

kJ/kg 61.68K)kJ/kg K)(1.3072 15.373(kJ/kg) 17.419(

−=⋅−=−=

−=⋅−=−=

ggg

fff

Tshg

Tshg

which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

16-69 It is to be shown that a mixture of saturated liquid water and saturated water vapor at 300 kPa satisfies the criterion for phase equilibrium.

Analysis The saturation temperature at 300 kPa is 406.7 K. Using the definition of Gibbs function and enthalpy and entropy data from Table A-5,

kJ/kg 6.118K)kJ/kg K)(6.9917 7.406(kJ/kg) 9.2724(

kJ/kg 5.118K)kJ/kg K)(1.6717 7.406(kJ/kg) 43.561(

−=⋅−=−=

−=⋅−=−=

ggg

fff

Tshg

Tshg

which are practically same. Therefore, the criterion for phase equilibrium is satisfied.

16-70 It is to be shown that a saturated liquid-vapor mixture of refrigerant-134a at -10°C satisfies the criterion for phase equilibrium.

Analysis Using the definition of Gibbs function and enthalpy and entropy data from Table A-11,

kJ/kg 235.2K)kJ/kg K)(0.93766 15.263(kJ/kg) 51.244(

kJ/kg 249.2K)kJ/kg K)(0.15504 15.263(kJ/kg) 55.38(

−=⋅−=−=

−=⋅−=−=

ggg

fff

Tshg

Tshg

which are sufficiently close. Therefore, the criterion for phase equilibrium is satisfied.

16-71 The number of independent properties needed to fix the state of a mixture of oxygen and nitrogen in the gas phase is to be determined.

Analysis In this case the number of components is C = 2 and the number of phases is PH = 1. Then the number of independent variables is determined from the phase rule to be

IV = C - PH + 2 = 2 - 1 + 2 = 3

Therefore, three independent properties need to be specified to fix the state. They can be temperature, the pressure, and the mole fraction of one of the gases.


16-64

16-72 The values of the Gibbs function for saturated refrigerant-134a at 0°C as a saturated liquid, saturated vapor, and a mixture of liquid and vapor are to be calculated.

Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is,

kJ/kg 3.97−=⋅−=−= )KkJ/kg 20439.0)(K 15.273(kJ/kg 86.51fff Tshg

For the vapor phase,

kJ/kg 3.96−=⋅−=−= )KkJ/kg 93139.0)(K 15.273(kJ/kg 45.250ggg Tshg

For the saturated mixture with a quality of 30%,

kJ/kg 3.96−=⋅−=−=

⋅=⋅+⋅=+=

=+=+=

)KkJ/kg 42249.0)(kJ/kg 15.273(kJ/kg 44.111

KkJ/kg 42249.0)KkJ/kg 2701(0.30)(0.7KkJ/kg 20439.0

kJ/kg 44.111)kJ/kg 60.198(0.30)(kJ/kg 86.51

Tshg

xsss

xhhh

fgf

fgf

The results agree and demonstrate that phase equilibrium exists.

16-73 The values of the Gibbs function for saturated refrigerant-134a at -10°C are to be calculated.

Analysis Obtaining properties from Table A-11, the Gibbs function for the liquid phase is,

kJ/kg 2.25−=⋅−=−= )KkJ/kg 15504.0)(K 15.263(kJ/kg 55.38fff Tshg

For the vapor phase,

kJ/kg 2.24−=⋅−=−= )KkJ/kg 93766.0)(K 15.263(kJ/kg 51.244ggg Tshg

The results agree and demonstrate that phase equilibrium exists.

16-74 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the nitrogen and the mass fraction of the oxygen at this temperature is to be determined.

Analysis From the equilibrium diagram (Fig. 16-21) we read T = 88 K.

For the liquid phase, from the same figure,

90.0O2, =fy and 10.0N2, =fy

Then,

0.911=+

=+

==)28)(10.0()32)(90.0(

)32)(90.0(mf

N2N2,O2O2,

O2O2,

total,

O2,O2, MyMy

Mymm

ff

f

f

ff

R-134a0°C

R-134a −10°C x = 0.4


16-65

16-75 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The pressure of ammonia is to be determined for two compositions of the liquid phase.

Assumptions The mixture is ideal and thus Raoult’s law is applicable.

Analysis According to Raoults’s law, when the mole fraction of the ammonia liquid is 20%,

kPa 123.1=== kPa) 3.615(20.0)(NH3sat,NH3,NH3 TPyP f

When the mole fraction of the ammonia liquid is 80%,

kPa 492.2=== kPa) 3.615(80.0)(NH3sat,NH3,NH3 TPyP f

16-76 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the liquid phase is given. The composition of the vapor phase is to be determined.


Properties At kPa 5.1003 and kPa 170.3 C,2532 NHsat,OHsat, ==° PP .

Analysis The vapor pressures are

kPa 74.501kPa) 5.1003(50.0)(

kPa 585.1kPa) 170.3(50.0)(

333

222

NHsat,NH,NH

OHsat,OH,OH

===

===

TPyP

TPyP

f

f

Thus the total pressure of the mixture is

kPa 33.503kPa )74.501585.1(32 NHOHtotal =+=+= PPP

Then the mole fractions in the vapor phase become

99.69%or

kPa 503.33kPa 74.501

0.31%or kPa 503.33

kPa 585.1

total

NHNH,

total

OHOH,

3

3

2

2

0.9969

0.0031

===

===

PP

y

PP

y

g

g

H2O + NH3 25°C

H2O + NH3 10°C


16-66

16-77 A liquid-vapor mixture of ammonia and water in equilibrium at a specified temperature is considered. The composition of the vapor phase is given. The composition of the liquid phase is to be determined.


Properties At kPa. 5.2033 and kPa 352.12 C,5032 NHsat,OHsat, ==° PP

Analysis We have %1OH, 2=gy and %99

3NH, =gy . For an ideal two-phase mixture we have

1

)(

)(

32

333

222

NH,OH,

NHsat,NH,NH,

OHsat,OH,OH,

=+

=

=

ff

fmg

fmg

yy

TPyPy

TPyPy

Solving for y f ,H O,2

)1(kPa) 352.12)(99.0(kPa) 5.2033)(01.0()1( OH,OH,

OHsat,NH,

NHsat,OH,OH, 22

23

32

2 ffg

gf yy

Py

Pyy −=−=

It yields

0.3760.624 ==32 NH,OH, and ff yy

16-78 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture, the composition of each phase at a specified temperature and pressure is to be determined.

Analysis From the equilibrium diagram (Fig. 16-21) we read

Liquid: 22 O 70% and N 30%

Vapor: 22 O 34% and N 66%

16-79 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the vapor phase.


16-80 Using the liquid-vapor equilibrium diagram of an oxygen-nitrogen mixture at a specified pressure, the temperature is to be determined for a specified composition of the liquid phase.


H2O + NH3 50°C


16-67

16-81 A rubber wall separates O2 and N2 gases. The molar concentrations of O2 and N2 in the wall are to be determined.

Assumptions The O2 and N2 gases are in phase equilibrium with the rubber wall.

Properties The molar mass of oxygen and nitrogen are 32.0 and 28.0 kg/kmol, respectively (Table A-1). The solubility of oxygen and nitrogen in rubber at 298 K are 0.00312 and 0.00156 kmol/m3⋅bar, respectively (Table 16-3).

Analysis Noting that 500 kPa = 5 bar, the molar densities of oxygen and nitrogen in the rubber wall are determined to be

3kmol/m 0.0156=

bar) 5)(bar.kmol/m 00312.0(

S)0(3

side gas ,Oside solid ,O 22

=

×= PC

C PN , solid side N , gas side

32 2

kmol / m bar bar)

=

( )

( . . )(

0

0 00156 5

= ×

=

S

0.0078 kmol / m3

That is, there will be 0.0156 kmol of O2 and 0.0078 kmol of N2 gas in each m3 volume of the rubber wall.

O2 25°C

500 kPa

N2 25°C

500 kPa

CO2 CN2

Rubber plate


16-68

16-82 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined.


Properties At kPa 10.10 C,46at and kPa 6112.0 C,0 H2Osat,H2Osat, =°=° PP (Table A-4). The saturation pressures of ammonia at the same temperatures are given to be 430.6 kPa and 1830.2 kPa, respectively.

Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by

H2Osat,NH3,H2Osat,H2O,H2Og,

NH3sat,NH3,NH3g,

)1( PyPyP

PyP

ff

f

−==

=

Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is

0.03294=⎯→⎯

−+=

−+=

NH3,NH3,NH3,

NH3,

H2Osat,NH3,NH3sat,NH3,

NH3sat,NH3,NH3,

)1(6112.06.4306.430

96.0

)1(

fff

f

ff

fg

yyy

y

PyPyPy

y

Then,

kPa 14.78=−+=

−+=

)6112.0)(03294.01()6.430)(03294.0(

)1( H2Osat,NH3,NH3sat,NH3, PyPyP ff

Performing the similar calculations for the regenerator,

0.1170=⎯→⎯−+

= NH3,NH3,NH3,

NH3,

)1(10.102.18302.1830

96.0 fff

f yyy

y

kPa 223.1=−+= )10.10)(1170.01()2.1830)(1170.0(P


16-69

16-83 An ammonia-water absorption refrigeration unit is considered. The operating pressures in the generator and absorber, and the mole fractions of the ammonia in the strong liquid mixture being pumped from the absorber and the weak liquid solution being drained from the generator are to be determined.


Properties At kPa 3851.7 C,40at and kPa 9353.0 C,6 H2Osat,H2Osat, =°=° PP (Table A-4 or EES). The saturation pressures of ammonia at the same temperatures are given to be 534.8 kPa and 1556.7 kPa, respectively.

Analysis According to Raoults’s law, the partial pressures of ammonia and water are given by

H2Osat,NH3,H2Osat,H2O,H2Og,

NH3sat,NH3,NH3g,

)1( PyPyP

PyP

ff

f

−==

=

Using Dalton’s partial pressure model for ideal gas mixtures, the mole fraction of the ammonia in the vapor mixture is

0.04028=⎯→⎯

−+=

−+=

NH3,NH3,NH3,

NH3,

H2Osat,NH3,NH3sat,NH3,

NH3sat,NH3,NH3,

)1(9353.08.5348.534

96.0

)1(

fff

f

ff

fg

yyy

y

PyPyPy

y

Then,

kPa 22.44=−+=

−+=

)9353.0)(04028.01()8.534)(04028.0(

)1( H2Osat,NH3,NH3sat,NH3, PyPyP ff

Performing the similar calculations for the regenerator,

0.1022=⎯→⎯−+

= NH3,NH3,NH3,

NH3,

)1(3851.77.15567.1556

96.0 fff

f yyy

y

kPa 165.7=−+= )3851.7)(1022.01()7.1556)(1022.0(P


16-70

16-84 A liquid mixture of water and R-134a is considered. The mole fraction of the water and R-134a vapor are to be determined.


Properties At kPa 07.572 and kPa 3392.2 C,20 Rsat,H2Osat, ==° PP (Tables A-4, A-11). The molar masses of water and R-134a are 18.015 and 102.03 kg/kmol, respectively (Table A-1).

Analysis The mole fraction of the water in the liquid mixture is

9808.0)03.102/1.0()015.18/9.0(

015.18/9.0

)/mf()/mf(/mf

RR,H2OH2O,

H2OH2O,

total

H2O,H2O,

=+

=

+==

MMM

NN

yff

fff

According to Raoults’s law, the partial pressures of R-134a and water in the vapor mixture are

kPa 10.98kPa) 07.572)(9808.01(Rsat,R,R, =−== PyP fg

kPa 2.294kPa) 3392.2)(9808.0(H2Osat,H2O,H2O, === PyP fg

The total pressure of the vapor mixture is then

kPa 274.13294.298.10H2O,R,total =+=+= gg PPP

Based on Dalton’s partial pressure model for ideal gases, the mole fractions in the vapor phase are

0.1728===kPa 13.274kPa 294.2

total

H2O,H2O, P

Py g

g

0.8272===kPa 13.274

kPa 98.10

total

R,R, P

Py g

g

H2O + R-134a 20°C


16-71

16-85E A mixture of water and R-134a is considered. The mole fractions of the R-134a in the liquid and vapor phases are to be determined.


Properties At psia 56.96 and psia 4597.0 F,77 Rsat,H2Osat, ==° PP (Tables A-4E, A-11E or EES).

Analysis According to Raoults’s law, the partial pressures of R-134a and water in the vapor phase are given by

psia) 4597.0(

psia) 56.96(

H2O,H2O,

H2O,H2Osat,H2O,H2O,

H2O,R,

R,Rsat,R,R,

ff

ffg

ff

ffg

NNN

PyP

NNN

PyP

+==

+==

The sum of these two partial pressures must equal the total pressure of the vapor mixture. In terms of

R,

H2O,

f

f

NN

x = , this sum is

7.141

4597.01

56.96=

++

+ xx

x

Solving this expression for x gives

x = 5.748 kmol H2O/kmol R-134a

In the vapor phase, the partial pressure of the R-134a vapor is

psia 31.141748.5

56.961

56.96R, =

+=

+=

xPg

The mole fraction of R-134a in the vapor phase is then

0.9735===psia 7.14psia 31.14R,

R, PP

y gg

According to Raoult’s law,

0.1482===psia 56.96psia 31.14

Rsat,

R,R, P

Py g

f

H2O + R-134a 14.7 psia, 77°F


16-72

16-86 A glass of water is left in a room. The mole fraction of the water vapor in the air and the mole fraction of air in the water are to be determined when the water and the air are in thermal and phase equilibrium.

Assumptions 1 Both the air and water vapor are ideal gases. 2 Air is saturated since the humidity is 100 percent. 3 Air is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for air dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 74,000 bar. Molar masses of dry air and water are 29 and 18 kg/kmol, respectively (Table A-1).

Analysis (a) Noting that air is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,

kPa 600.3C27 @sat vapor == °PP (Table A-4)

Assuming both the air and vapor to be ideal gases, the mole fraction of water vapor in the air is determined to be

0.0371===kPa 97

kPa 600.3vaporvapor P

Py

(b) Noting that the total pressure is 97 kPa, the partial pressure of dry air is

bar 0.934=kPa 4.93600.397vaporairdry =−=−= PPP

From Henry’s law, the mole fraction of air in the water is determined to be

5101.26 −×===bar 74,000bar 934.0side gasair,dry

sideliquidair,dry HP

y

Discussion The amount of air dissolved in water is very small, as expected.

Water 27ºC

Air 27ºC

97 kPa φ = 100%


16-73

16-87 A carbonated drink in a bottle is considered. Assuming the gas space above the liquid consists of a saturated mixture of CO2 and water vapor and treating the drink as a water, determine the mole fraction of the water vapor in the CO2 gas and the mass of dissolved CO2 in a 300 ml drink are to be determined when the water and the CO2 gas are in thermal and phase equilibrium.

Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 and the water vapor are ideal gases. 3 The CO2 gas and water vapor in the bottle from a saturated mixture. 4 The CO2 is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 27°C is 3.568 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 27ºC (300 K) is given in Table 16-2 to be H = 1710 bar. Molar masses of CO2 and water are 44 and 18 kg/kmol, respectively (Table A-1).

Analysis (a) Noting that the CO2 gas in the bottle is saturated, the partial pressure of water vapor in the air will simply be the saturation pressure of water at 27°C,

kPa 568.3C27 @sat vapor == °PP (more accurate EES value compared to interpolation value from Table A-4)

Assuming both CO2 and vapor to be ideal gases, the mole fraction of water vapor in the CO2 gas becomes

0.0274===kPa 130kPa 568.3vapor

vapor PP

y

(b) Noting that the total pressure is 130 kPa, the partial pressure of CO2 is

bar 1.264=kPa 4.126568.3130vaporgas CO2=−=−= PPP

From Henry’s law, the mole fraction of CO2 in the drink is determined to be

yP

HCO ,liquid sideCO ,gas side

2

2 bar1710 bar

= = = × −1264.7.39 10 4

Then the mole fraction of water in the drink becomes

y ywater, liquid side CO , liquid side2= − = − × =−1 1 7 39 10 0 99934. .

The mass and mole fractions of a mixture are related to each other by

m

ii

mm

ii

m

ii M

My

MNMN

mm

===mf

where the apparent molar mass of the drink (liquid water - CO2 mixture) is

M y M y M y Mm i i= = + = × + × × =−∑ liquid water water CO CO2 2

kg / kmol0 9993 18 0 7 39 10 44 18 024. . ( . ) .

Then the mass fraction of dissolved CO2 gas in liquid water becomes

0.0018002.18

441039.7)0(mf 4CO

sideliquid,COsideliquid,CO2

22=×== −

mM

My

Therefore, the mass of dissolved CO2 in a 300 ml ≈ 300 g drink is

g 0.54=== g) 300)(00180.0(mf22 COCO mmm


16-74

Review Problems

16-88 The equilibrium constant of the dissociation process O2 ↔ 2O is given in Table A-28 at different temperatures. The value at a given temperature is to be verified using Gibbs function data.

Analysis The KP value of a reaction at a specified temperature can be determined from the Gibbs function data using


p uu= = −−Δ Δ*( )/ ln *( ) / or

where

kJ/kmol 375,243)655.26820008682881,670(1

)135.20120006852564,42190,249(2

])[(])[(

)()(

)()()(*

22

22

22

O2982000OO2982000O

OOOO

*OO

*OO

=×−−+×−

×−−+×=

−−+−−−+=

−−−=

−=Δ

sThhhsThhh

sThsTh

TgTgTG

ff νν

νν

νν

Substituting,

636.14K)] K)(2000kJ/kmol (8.314kJ/kmol)/[ 375,243(ln −=⋅−=pK

or

K p = × −4.4 10 7 (Table A-28: ln KP = -14.622)

16-89 A mixture of H2 and Ar is heated is heated until 15% of H2 is dissociated. The final temperature of mixture is to be determined.

Assumptions 1 The constituents of the mixture are ideal gases. 2 Ar in the mixture remains an inert gas.


Stoichiometric: )2 and 1 (thus 2HH HH2 2==⇔ νν

Actual: { {inertreact.

2prod

2 ArH85.0H3.0Ar+H ++⎯→⎯43421

The equilibrium constant KP can be determined from

0492.013.085.0

185.03.0 122

totalH

H2HH

2H

2

H

=⎟⎠⎞

⎜⎝⎛

++=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−−νν

ν

ν

NP

N

NK p

From Table A-28, the temperature corresponding to this KP value is T = 3117 K.

O2 ↔ 2O

2000 K

H H2 2⇔ Ar

1 atm


16-75

16-90 A mixture of H2O, O2, and N2 is heated to a high temperature at a constant pressure. The equilibrium composition is to be determined.

Assumptions 1 The equilibrium composition consists of H2O, O2, N2 and H2. 2 The constituents of the mixture are ideal gases.


Stoichiometric: ) and ,1 ,1 (thus O+HOH 21

OHOH221

22 222===⇔ ννν

Actual: {inert

2products

22react.

2222 N5O+HOHN5O2+OH ++⎯→⎯+43421321zyx

H balance: 2 2 2 1= + ⎯→⎯ = −x y y x

O balance: xzzx 5.05.225 −=⎯→⎯+=

Total number of moles: xzyxN 5.05.85total −=+++=


15.01

total

5.0)(

totalOH

OHO2H2O2H

O2H

2

2O

22H

2

−+−−

⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

NP

xzy

NP

N

NNK p

ννν

ν

νν

From Table A-28, lnKP = -6.768 at 2200 K. Thus KP = 0.00115. Substituting,

5.05.0

5.05.85)5.05.1)(1(00115.0 ⎟

⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.9981

Then,

y = 1 - x = 0.0019

z = 2.5 - 0.5x = 2.00095


2222 5N2.00095O0.0019H+O0.9981H ++

The equilibrium constant for the reaction H O OH + H212⇔ 2 is lnKP = -7.148, which is very close to the

KP value of the reaction considered. Therefore, it is not realistic to assume that no OH will be present in equilibrium mixture.

1 H2O 2 O2 5 N2

2200 K 5 atm


16-76

CH4

25°C

Air

25°C

Combustion chamber

1 atm

CO CO2 H2O O2 N2

16-91 [Also solved by EES on enclosed CD] Methane gas is burned with stoichiometric amount of air during a combustion process. The equilibrium composition and the exit temperature are to be determined.

Assumptions 1 The product gases consist of CO2, H2O, CO, N2, and O2. 2 The constituents of the mixture are ideal gases. 3 This is an adiabatic and steady-flow combustion process.

Analysis (a) The combustion equation of CH4 with stoichiometric amount of O2 can be written as

CH O N CO + (0.5 0.5 )O H O + 7.52N4 2 2 2 2 2 2+ + ⎯→⎯ + − − +2 376 1 2( . ) ( )COx x x

After combustion, there will be no CH4 present in the combustion chamber, and H2O will act like an inert gas. The equilibrium equation among CO2, CO, and O2 can be expressed as

) and ,1 ,1 (thus O+COCO 21

OCOCO221

2 22===⇔ ννν

and

)(

totalCO

OCO2CO2OCO

2CO

2

2O

2CO ννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

where

N x x x xtotal = + − + − + + = −( ) ( . . ) . . .1 15 0 5 2 7 52 12 02 0 5

Substituting, 15.15.0

5.002.121)5.05.0)(1( −

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xxK p

The value of KP depends on temperature of the products, which is yet to be determined. A second relation to determine KP and x is obtained from the steady-flow energy balance expressed as

( ) ( ) ( ) ∑∑∑∑ −−+=⎯→⎯−+−−+=RfRPfPRfRPfP hNhhhNhhhNhhhN ooooooo 00

since the combustion is adiabatic and the reactants enter the combustion chamber at 25°C. Assuming the air and the combustion products to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

CH4(g) -74,850 --

N2 0 8669

O2 0 8682

H2O(g) -241,820 9904

CO -110,530 8669

CO2 -393,520 9364

Substituting,

0 393 520 9364 1 110 530 8669

2 241 820 9904 0 5 0 5 0 8682

7 52 0 8669 1 74 850 0 0298 298

= − + − + − − + −

+ − + − + − + −

+ + − − − + − − −

x h x h

h x h

h h h

( , ) ( )( , )

( , ) ( . . )( )

. ( ) ( , )

CO CO

H O O

N

2

2 2

2


16-77

which yields

xh x h h x h h xCO CO H O O N2 2 2 2+ − + + − + − =( ) ( . . ) . , ,1 2 0 5 0 5 7 52 279 344 617 329

Now we have two equations with two unknowns, TP and x. The solution is obtained by trial and error by assuming a temperature TP, calculating the equilibrium composition from the first equation, and then checking to see if the second equation is satisfied. A first guess is obtained by assuming there is no CO in the products, i.e., x = 1. It yields TP = 2328 K. The adiabatic combustion temperature with incomplete combustion will be less.

Take K

Take K

T K x RHS

T K x RHS

p p

p p

= ⎯→⎯ = − ⎯→⎯ = ⎯→⎯ =

= ⎯→⎯ = − ⎯→⎯ = ⎯→⎯ =

2300 4 49 0 870 641 093

2250 4 805 0 893 612 755

ln . . ,

ln . . ,

By interpolation,

T xp = =2258 K and 0889.

Thus the composition of the equilibrium mixture is

0.889CO + 0.111CO 0.0555O 2H O 7.52N2 2 2 2+ + +


16-78

16-92 EES Problem 16-91 is reconsidered. The effect of excess air on the equilibrium composition and the exit temperature by varying the percent excess air from 0 to 200 percent is to be studied.

Analysis The problem is solved using EES, and the solution is given below. "Often, for nonlinear problems such as this one, good gusses are required to start the solution. First, run the program with zero percent excess air to determine the net heat transfer as a function of T_prod. Just press F3 or click on the Solve Table icon. From Plot Window 1, where Q_net is plotted vs T_prod, determnine the value of T_prod for Q_net=0 by holding down the Shift key and move the cross hairs by moving the mouse. Q_net is approximately zero at T_prod = 2269 K. From Plot Window 2 at T_prod = 2269 K, a, b, and c are approximately 0.89, 0.10, and 0.056, respectively." "For EES to calculate a, b, c, and T_prod directly for the adiabatic case, remove the '{ }' in the last line of this window to set Q_net = 0.0. Then from the Options menu select Variable Info and set the Guess Values of a, b, c, and T_prod to the guess values selected from the Plot Windows. Then press F2 or click on the Calculator icon." "Input Data" {PercentEx = 0} Ex = PercentEX/100 P_prod =101.3 [kPa] R_u=8.314 [kJ/kmol-K] T_fuel=298 [K] T_air=298 [K] "The combustion equation of CH4 with stoichiometric amount of air is CH4 + (1+Ex)(2)(O2 + 3.76N2)=CO2 +2H2O+(1+Ex)(2)(3.76)N2" "For the incomplete combustion process in this problem, the combustion equation is CH4 + (1+Ex)(2)(O2 + 3.76N2)=aCO2 +bCO + cO2+2H2O+(1+Ex)(2)(3.76)N2" "Specie balance equations" "O" 4=a *2+b +c *2+2 "C" 1=a +b N_tot =a +b +c +2+(1+Ex)*(2)*3.76 "Total kilomoles of products at equilibrium" "We assume the equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 16-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 16-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )=K_P *a "Conservation of energy for the reaction, assuming SSSF, neglecting work , ke, and pe:" E_in - E_out = DELTAE_cv E_in = Q_net + HR "The enthalpy of the reactant gases is" HR=enthalpy(CH4,T=T_fuel)+ (1+Ex)*(2) *enthalpy(O2,T=T_air)+(1+Ex)*(2)*3.76 *enthalpy(N2,T=T_air) E_out = HP "The enthalpy of the product gases is"


16-79

HP=a *enthalpy(CO2,T=T_prod )+b *enthalpy(CO,T=T_prod ) +2*enthalpy(H2O,T=T_prod )+(1+Ex)*(2)*3.76*enthalpy(N2,T=T_prod ) + c *enthalpy(O2,T=T_prod ) DELTAE_cv = 0 "Steady-flow requirement" Q_net=0 "For an adiabatic reaction the net heat added is zero."

PercentEx Tprod [K] 0 2260

20 2091 40 1940 60 1809 80 1695

100 1597 120 1511 140 1437 160 1370 180 1312 200 1259

0 40 80 120 160 2001200

1400

1600

1800

2000

2200

2400

Percent Excess Air [%]

T pro

d [K

]

1200 1400 1600 1800 2000 2200 2400 2600-0.10

0.10

0.30

0.50

0.70

0.90

1.10

Tprod, K

Coe

ffici

ents

: a, b

, c

Coefficients for CO2, CO, and O2 vs Tprod

a CO2a CO2b COb COc O2c O2


16-80

16-93 The equilibrium constant for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 100 kPa and 2000 K is to be determined.

Assumptions 1 The constituents of the mixture are ideal gases.

Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide,

24 2HC CH +⇔ 847.7−= eK P

22 CO OC ⇔+ 839.23eK P =

When these two reactions are summed and the common carbon term cancelled, the result is

2224 2HCOO CH +⇔+ 992.15)847.7839.23( eeK P == −

Next, we include the water dissociation reaction (Table A-28),

OH2O 2H 222 ⇔+ 29.16)145.8(2 eeK P ==

which when summed with the previous reaction and the common hydrogen term is cancelled yields

O2HCOO2 CH 2224 +⇔+ 282.3229.16992.15 eeK P == +

Then,

32.282=PKln

CH4+2O2 ⇔ CO2+2H2O

2000 K 100 kPa


16-81

16-94 The equilibrium mole fraction of the water vapor for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 100 kPa and 2000 K is to be determined. Assumptions 1 The equilibrium composition consists of CH4, O2, CO2, and H2O. 2 The constituents of the mixture are ideal gases. Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide, 24 2HC CH +⇔ 847.7−= eK P

22 CO OC ⇔+ 839.23eK P = When these two reactions are summed and the common carbon term cancelled, the result is 2224 2HCOO CH +⇔+ 992.15)847.7839.23( eeK P == − Next, we include the water dissociation reaction, OH2O 2H 222 ⇔+ 29.16)145.8(2 eeK P == which when summed with the previous reaction and the common hydrogen term is cancelled yields O2HCOO2 CH 2224 +⇔+ 282.3229.16992.15 eeK P == + Then, 28232ln .K P =

Actual reeaction: 44 344 214434421

products22

react.2424 OH+COOCHO2CH mzyx ++⎯→⎯+

C balance: xzzx −=⎯→⎯+= 11

H balance: xmmx 22244 −=⎯→⎯+=

O balance: xymzy 2224 =⎯→⎯++=

Total number of moles: 3total =+++= mzyxN The equilibrium constant relation can be expressed as

O2CH4H2OCO2

O2CH4

H2OCO2

totalO2CH4

H2OCO2νννν

νν

νν −−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NNK p

Substituting,

2121

2

2282.32

3325.101/100

)2()22)(1( −−+

⎟⎠⎞

⎜⎝⎛−−

=xx

xxe

Solving for x, x = 0.00002122 Then, y = 2x = 0.00004244 z = 1 − x = 0.99997878 m = 2 − 2x = 1.99995756 Therefore, the equilibrium composition of the mixture at 2000 K and 100 kPa is OH 1.99995756CO 0.99997878O 0.00004244+CH 0.00002122 2224 ++ The mole fraction of water vapor is then

0.6667==3

99995756.1H2Oy

CH4+2O2 ⇔ CO2+2H2O

2000 K 100 kPa


16-82

16-95 The equilibrium partial pressure of the carbon dioxide for the reaction CH4 + 2O2 ⇔ CO2 + 2H2O at 700 kPa and 3000 K is to be determined. Assumptions 1 The equilibrium composition consists of CH4, O2, CO2, and H2O. 2 The constituents of the mixture are ideal gases. Analysis This is a simultaneous reaction. We can begin with the dissociation of methane and carbon dioxide, 24 2HC CH +⇔ 685.9−= eK P

22 CO OC ⇔+ 869.15eK P = When these two reactions are summed and the common carbon term cancelled, the result is 2224 2HCOO CH +⇔+ 184.6)685.9869.15( eeK P == − Next, we include the water dissociation reaction, OH2O 2H 222 ⇔+ 172.6)086.3(2 eeK P == which when summed with the previous reaction and the common hydrogen term is cancelled yields O2HCOO2 CH 2224 +⇔+ 356.12172.6184.6 eeK P == + Then, 356.12ln =PK

Actual reeaction: 44 344 214434421

products22

react.2424 OH+COOCHO2CH mzyx ++⎯→⎯+

C balance: xzzx −=⎯→⎯+= 11

H balance: xmmx 22244 −=⎯→⎯+=

O balance: xymzy 2224 =⎯→⎯++= Total number of moles: 3total =+++= mzyxN The equilibrium constant relation can be expressed as

O2CH4H2OCO2

O2CH4

H2OCO2

totalO2CH4

H2OCO2νννν

νν

νν −−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NNK p

Substituting,

2121

2

2356.12

3325.101/700

)2()22)(1( −−+

⎟⎠⎞

⎜⎝⎛−−

=xx

xxe

Solving for x, x = 0.01601 Then, y = 2x = 0.03202 z = 1 − x = 0.98399 m = 2 − 2x = 1.96798 Therefore, the equilibrium composition of the mixture at 3000 K and 700 kPa is OH 1.96798CO 0.98399O 0.03202+CH 0.01601 2224 ++ The mole fraction of carbon dioxide is

0.32803

98399.0CO2 ==y

and the partial pressure of the carbon dioxide in the product mixture is kPa 230=== )kPa 700)(3280.0(CO2CO2 PyP

CH4+2O2 ⇔ CO2+2H2O

3000 K 700 kPa


16-83

16-96 Methane is heated from a specified state to another state. The amount of heat required is to be determined without and with dissociation cases. Properties The molar mass and gas constant of methane are 16.043 kg/kmol and 0.5182 kJ/kg⋅K (Table A-1).


Analysis (a) An energy balance for the process gives

[ ])(

)(

1212

12in


system


outin

TTRhhN

uuNQ

EEE

u −−−=

−=

Δ=−4342143421

Using the empirical coefficients of Table A-2c,

kJ/kmol 239,38

)2981000(4

1001.11

)2981000(3

10269.1)2981000(2

05024.0)2981000(89.19

)(4

)(3

)(2

)(

449

335

22

41

42

31

32

21

2212

2

112

=

−×−

+

−×

+−+−=

−+−+−+−==−

−

−

∫ TTdTTcTTbTTadTchh p

Substituting,

[ ] kJ 324,000=−⋅−= 298)KK)(1000kJ/kmol (8.314kJ/kmol 38,239kmol) 10(inQ


Stoichiometric: 24 2HC CH +⇔ )2 and 1 ,1 (thus H2CCH4 === ννν

Actual: 43421321

products2

react.44 HCCHCH zyx ++⎯→⎯


H balance: xzzx 22244 −=⎯→⎯+=

Total number of moles: xzyxN 23total −=++=


CH4H2C

CH4

H2C

totalCH4

H2Cννν

ν

νν −+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK p

From the problem statement, at 1000 K, 328.2ln −=pK . Then,

09749.0328.2 == −eK P

Substituting,

CH4

1000 K 1 atm


16-84

1212

231)22)(1(09749.0

−+

⎟⎠⎞

⎜⎝⎛

−−−

=xx

xx

Solving for x,

x = 0.6414

Then,

y = 1 − x = 0.3586

z = 2 − 2x = 0.7172


24 H 0.7172C 3586.0CH 0.6414 ++

The mole fractions are

0.41777172.17172.0

0.20887172.13586.0

0.37357172.16414.0

7172.03586.06414.06414.0

total

H2H2

total

CC

total

CH4CH4

===

===

==++

==

NN

y

NN

y

NN

y

The heat transfer can be determined from

[ ]kJ 245,700=

−++=

−++=

)298)(8.27)(10()1000)(711.0)(2088.0()1000)(7.21)(4177.0()1000)(3.63)(3735.0()10()( 1CH4,2C,C2H2,H22CH4,CH4in TNcTcyTcyTcyNQ vvvv


16-85

16-97 Solid carbon is burned with a stoichiometric amount of air. The number of moles of CO2 formed per mole of carbon is to be determined.


Analysis Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by

2222 N76.3CO)3.76N(OC +⎯→⎯++

The number of moles of CO2 in the products is then

1=C

CO2

NN

16-98 Solid carbon is burned with a stoichiometric amount of air. The amount of heat released per kilogram of carbon is to be determined.


Analysis Inspection of Table A-28 reveals that the dissociation equilibrium constants of CO2, O2, and N2 are quite small and therefore may be neglected. (We learned from another source that the equilibrium constant for CO is also small). The combustion is then complete and the reaction is described by

2222 N76.3CO)3.76N(OC +⎯→⎯++

The heat transfer for this combustion process is determined from the energy balance systemoutin EEE Δ=− applied on the combustion chamber with W = 0. It reduces to


out

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

Substance

ofh

kJ/kmol

K298h

kJ/kmol

K1000h

kJ/kmol

N2 0 8669 30,129

CO2 -393,520 9364 42,769

Substituting,

( ) ( )C kJ/kmol 279,400

8669129,300)76.3(9364769,42520,393)1(out

−=−++−+−=−Q

or C kJ/kg 23,280==kg/kmol 12

kJ/kmol 279,400outQ

Carbon + Air 25°C

Carbon + Air 25°C


16-86

16-99 Methane gas is burned with 30 percent excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases. Analysis Inspection of the equilibrium constants of the possible reactions indicate that only the formation of NO need to be considered in addition to other complete combustion products. Then, the stoichiometric and actual reactions in this case are Stoichiometric: )2 and ,1 ,1 (thus NO2ON NOO2N222 ===⇔+ ννν

Actual: 2222224 NONOOH2CO)N76.3O(6.2CH zyx ++++⎯→⎯++

N balance: xzzx 5.0888.42776.9 −=⎯→⎯+= O balance: xyyx 5.06.02222.5 −=⎯→⎯+++= Total number of moles: 488.821total =++++= zyxN The equilibrium constant relation can be expressed as

)(

totalO2N2

NOO2N2NO

O2N2

NO ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, at 1600 K, 294.5ln −=pK . Since the stoichiometric reaction being considered is double this reaction, 510522.2)294.52exp( −×=×−=pK Substituting,

1122

5

488.81

)5.0888.4)(5.06.0(10522.2

−−− ⎟

⎠⎞

⎜⎝⎛

−−=×

xxx

Solving for x, x = 0.008566 Then, y = 0.6 − 0.5x = 0.5957 z = 4.888 − 0.5x =4.884 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is 2222224 4.884N0.5957O0.008566NOO2HCO)3.76N2.6(OCH ++++⎯→⎯++ The heat transfer for this combustion process is determined from the energy balance systemoutin EEE Δ=− applied on the combustion chamber with W = 0. It reduces to ( ) ( )∑ ∑ −+−−+=−

RfRPfP hhhNhhhNQ ooooout

Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables, Substance

ofh

kJ/kmol K298h

kJ/kmol K1600h

kJ/kmol CH4 -74,850 --- --- O2 0 8682 52,961 N2 0 8669 50,571 H2O -241,820 9904 62,748 CO2 -393,520 9364 76,944

Neglecting the effect of NO in the energy balance and substituting, ( )

4

out

CH kJ/kmol ,900472)850,74()8669571,50)(884.4(

)8682961,52(5957.0)9904748,62820,241)(2(9364944,76520,393)1(

−=−−−+

−+−+−+−+−=−Q

or 4CH kJ/kmol 472,900=outQ

CH4

25°C

30% excess air

25°C

Qout

Combustion chamber

1 atm

CO2, H2O NO, O2, N2

1600 K


16-87

16-100 Propane gas is burned with 30% excess air. The equilibrium composition of the products of combustion and the amount of heat released by this combustion are to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases. Analysis (a) The stoichiometric and actual reactions in this case are Stoichiometric: )2 and ,1 ,1 (thus NO2ON NOO2N222 ===⇔+ ννν

Actual: 22222283 NONOOH43CO)N76.3O(53.1HC zyx ++++⎯→⎯+×+

N balance: xzzx 5.044.24288.48 −=⎯→⎯+= O balance: xyyx 5.05.124613 −=⎯→⎯+++= Total number of moles: 94.3243total =++++= zyxN The equilibrium constant relation can be expressed as

)(

totalO2N2

NOO2N2NO

O2N2

NO ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, at 1600 K, 294.5ln −=pK . Since the stoichiometric reaction being considered is double this reaction, 510522.2)294.52exp( −×=×−=pK Substituting,

1122

5

94.321

)5.044.24)(5.05.1(10522.2

−−− ⎟

⎠⎞

⎜⎝⎛

−−=×

xxx

Solving for x, x = 0.03024 Then, y = 1.5 − 0.5x = 1.485 z = 24.44 − 0.5x = 24.19 Therefore, the equilibrium composition of the products mixture at 1600 K and 1 atm is 22222283 24.19N1.485O0.03024NOO4H3CO)3.76N6.5(OHC ++++⎯→⎯++ (b) The heat transfer for this combustion process is determined from the energy balance

systemoutin EEE Δ=− applied on the combustion chamber with W = 0. It reduces to


out Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,

Substance

ofh

kJ/kmol K298h

kJ/kmol K1600h

kJ/kmol C3H8 -103,850 --- --- O2 0 8682 52,961 N2 0 8669 50,571 H2O -241,820 9904 62,748 CO2 -393,520 9364 76,944

Neglecting the effect of NO in the energy balance and substituting,

( )

83

out

HC kJ/kmol 360,654)850,103()8669571,50)(19.24(

)8682961,52(485.1)9904748,62820,241)(4(9364944,76520,393)3(

−=−−−+

−+−+−+−+−=−Q

or 83HC kJ/kg 14,870==kg/kmol 44

kJ/kmol 360,654outQ

Products

1600 K

C3H8

25°C

30% excess air

25°C

Qout

Combustion chamber

1 atm


16-88

16-101E Gaseous octane gas is burned with 40% excess air. The equilibrium composition of the products of combustion is to be determined. Assumptions 1 The equilibrium composition consists of CO2, H2O, O2, NO, and N2. 2 The constituents of the mixture are ideal gases.


Stoichiometric: )2 and ,1 ,1 (thus NO2ON NOO2N222 ===⇔+ ννν

Actual: 222222188 NONOOH98CO)N76.3O(5.124.1HC zyx ++++⎯→⎯+×+

N balance: xzzx 5.08.6526.131 −=⎯→⎯+=

O balance: xyyx 5.05291635 −=⎯→⎯+++=

Total number of moles: 8.8798total =++++= zyxN


)(

totalO2N2

NOO2N2NO

O2N2

NO ννν

νν

ν −−

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

NN

NK p

From Table A-28, at 2000 K (3600 R), 931.3ln −=pK . Since the stoichiometric

reaction being considered is double this reaction,

410851.3)931.32exp( −×=×−=pK

Substituting,

1122

4

8.877.14/600

)5.08.65)(5.05(10851.3

−−− ⎟

⎠⎞

⎜⎝⎛

−−=×

xxx

Solving for x,

x = 0.1119

Then,

y = 5 − 0.5x = 4.944

z = 65.8 − 0.5x = 65.74

Therefore, the equilibrium composition of the products mixture at 3600 R and 600 psia is

222222188 65.74N4.944O0.1119NOO9H8CO)3.76N17.5(OHC ++++⎯→⎯++

C8H18

40% excess air

Combustion chamber

600 psia

CO2, H2O NO, O2, N2

3600 R


16-89

16-102 A mixture of H2O and O2 is heated to a high temperature. The equilibrium composition is to be determined.



2H O + 3O H O H O + OH2 2 2 2 2⎯→⎯ + +x y z w


H balance: wyx ++= 224 (1)

O balance: wzx ++= 28 (2)


H O H O2 212 2⇔ + (reaction 1)

H O H OH212 2⇔ + (reaction 2)


ln . .

ln . .

K K

K KP P

P P

1 1

2 2

1392 0 24858

1088 0 33689

= − ⎯ →⎯ =

= − ⎯ →⎯ =


)(

totalOH

OHH2

)(

totalOH

OH1

O2HOH2H

O2H

2

OH2H

2

O2H2O2H

O2H

2

2O

2

2H

2

ννν

ν

νν

ννν

ν

νν

−+

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NNK

NP

N

NNK

P

P

where

wzyxNNNNN +++=+++= OHOHOHtotal 222

Substituting,

2/12/1 8))((24858.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

zy (3)

2/12/1 8))((33689.0 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

yw (4)


x = 1.371 y = 0.1646 z = 2.85 w = 0.928

Therefore, the equilibrium composition becomes

1.371H O 0.165H 2.85O 0.928OH2 2 2+ + +

H2O, OH, H2, O2

3600 K 8 atm


16-90

16-103 A mixture of CO2 and O2 is heated to a high temperature. The equilibrium composition is to be determined.

Assumptions 1 The equilibrium composition consists of CO2, CO, O2, and O. 2 The constituents of the mixture are ideal gases.


3C O + 3O CO CO O O2 2 2 2⎯→⎯ + + +x y z w

Mass balances for carbon and oxygen yield

C balance: 3 = +x y (1)

O balance: wzyx +++= 2212 (2)


221

2 OCOCO +⇔ (reaction 1)



ln . .

ln . .

K K

K KP P

P P

1 1

2 2

0169 11841

1935 01444

= ⎯→⎯ =

= − ⎯→⎯ =


2OO

2O

2

O

2CO2OCO

2CO

2

2O

2

CO

totalO

O2

)(

totalCO

OCO1

νν

ν

ν

ννν

ν

νν

−

−+

⎟⎟⎠

⎞⎜⎜⎝

⎛=

⎟⎟⎠

⎞⎜⎜⎝

⎛=

NP

N

NK

NP

N

NNK

P

P

where

wzyxNNNNN +++=+++= OCOOCOtotal 22

Substituting,

2/12/1 2))((1841.1 ⎟⎟

⎠

⎞⎜⎜⎝

⎛+++

=wzyxx

zy (3)

122 21444.0−

⎟⎟⎠

⎞⎜⎜⎝

⎛+++

=wzyxz

w (4)


x = 1.313 y = 1.687 z = 3.187 w = 1.314

Thus the equilibrium composition is

1.313CO 1.687CO 3.187O 1.314O2 2+ + +

CO2, CO, O2, O 3400 K 2 atm


16-91

16-104 EES Problem 16-103 is reconsidered. The effect of pressure on the equilibrium composition by varying pressure from 1 atm to 10 atm is to be studied.


"For EES to calculate a, b, c, and d at T_prod and P_prod press F2 or click on the Calculator icon. The EES results using the built in function data is not the same as the anwers provided with the problem. However, if we supply the K_P's from Table A-28 to ESS, the results are equal to the answer provided. The plot of moles CO vs. P_atm was done with the EES property data." "Input Data" P_atm = 2 [atm] P_prod =P_atm*101.3 R_u=8.314 [kJ/kmol-K] T_prod=3400 [K] P=P_atm "For the incomplete combustion process in this problem, the combustion equation is 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO" "Specie balance equations" "O" 3*2+3*2=a *2+b +c *2+d*1 "C" 3*1=a*1 +b*1 N_tot =a +b +c +d "Total kilomoles of products at equilibrium" "We assume the equilibrium reactions are CO2=CO+0.5O2 O2=2O" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "EES does not have a built-in property function for monatomic oxygen so we will use the JANAF procedure, found under Options/Function Info/External Procedures. The units for the JANAF procedure are kmol, K, and kJ. The values are calculated for 1 atm. The entropy must be corrected for other pressrues." Call JANAF('O',T_prod:Cp,h_O,s_O) "Units from JANAF are SI" "The entropy from JANAF is for one atmosphere and that's what we need for this approach." g_O=h_O-T_prod*s_O "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_CO+0.5*g_O2-1*g_CO2 DELTAG_2 =2*g_O-1*g_O2 "The equilibrium constants are given by Eq. 15-14." {K_P_2=0.1444 "From Table A-28" K_P_1 = 0.8445}"From Table A-28" K_p_1 = exp(-DELTAG_1/(R_u*T_prod)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod)) "From EES data"


16-92

"The equilibrium constant is also given by Eq. 15-15." "Write the equilibrium constant for the following system of equations: 3 CO2 + 3 O2=aCO2 +bCO + cO2+dO CO2=CO+0.5O2 O2=2O" "K_ P_1 = (P/N_tot)^(1+0.5-1)*(b^1*c^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(c )/a=K_P_1 "K_ P_2 = (P/N_tot)^(2-1)*(d^2)/(c^1)" P/N_tot *d^2/c =K_P_2

b [kmolCO] Patm [atm] 1.968 1 1.687 2 1.52 3

1.404 4 1.315 5 1.244 6 1.186 7 1.136 8 1.093 9 1.055 10

1 2 3 4 5 6 7 8 9 101

1.2

1.4

1.6

1.8

2

Patm [atm]

b [k

mol

CO

]


16-93

16-105 The hR at a specified temperature is to be determined using enthalpy and Kp data.


Analysis (a) The complete combustion equation of H2 can be expressed as

OHO+H 2221

2 ⇔

The hR of the combustion process of H at 2400 K2 is the amount of energy released as one kmol of H2 is burned in a steady-flow combustion chamber at a temperature of 2400 K, and can be determined from


Assuming the H2O, H2, and O2 to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2400 K

kJ/kmol

H2O -241,820 9904 103,508

H2 0 8468 75,383

O2 0 8682 83,174

Substituting,

hR = − + −− + −− + −

= −

1 241820 103 508 99041 0 75 383 846805 0 83174 8682

( , , )( , ). ( , )

252,377 kJ / kmol


⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

kJ/kmol -252,047≅

⎟⎠⎞

⎜⎝⎛ −

⋅≅−

R

R

h

hK 2600

1K 2200

1KkJ/kmol 314.8

768.6648.4


16-94

16-106 EES Problem 16-105 is reconsidered. The effect of temperature on the enthalpy of reaction using both methods by varying the temperature from 2000 to 3000 K is to be investigated.


"Input Data" T_prod=2400 [K] DELTAT_prod =25 [K] R_u=8.314 [kJ/kmol-K] T_prod_1 = T_prod - DELTAT_prod T_prod_2 = T_prod + DELTAT_prod "The combustion equation is 1 H2 + 0.5 O2 =>1 H2O" "The enthalpy of reaction H_bar_R using enthalpy data is:" h_bar_R_Enthalpy = HP - HR HP = 1*Enthalpy(H2O,T=T_prod ) HR = 1*Enthalpy(H2,T=T_prod ) + 0.5*Enthalpy(O2,T=T_prod ) "The enthalpy of reaction H_bar_R using enthalpy data is found using the following equilibruim data:" "The following equations provide the specific Gibbs function (g=h-Ts) for each component as a function of its temperature at 1 atm pressure, 101.3 kPa" g_H2O_1=Enthalpy(H2O,T=T_prod_1 )-T_prod_1 *Entropy(H2O,T=T_prod_1 ,P=101.3) g_H2_1=Enthalpy(H2,T=T_prod_1 )-T_prod_1 *Entropy(H2,T=T_prod_1 ,P=101.3) g_O2_1=Enthalpy(O2,T=T_prod_1 )-T_prod_1 *Entropy(O2,T=T_prod_1 ,P=101.3) g_H2O_2=Enthalpy(H2O,T=T_prod_2 )-T_prod_2 *Entropy(H2O,T=T_prod_2 ,P=101.3) g_H2_2=Enthalpy(H2,T=T_prod_2 )-T_prod_2 *Entropy(H2,T=T_prod_2 ,P=101.3) g_O2_2=Enthalpy(O2,T=T_prod_2 )-T_prod_2 *Entropy(O2,T=T_prod_2 ,P=101.3) "The standard-state (at 1 atm) Gibbs functions are" DELTAG_1 =1*g_H2O_1-0.5*g_O2_1-1*g_H2_1 DELTAG_2 =1*g_H2O_2-0.5*g_O2_2-1*g_H2_2 "The equilibrium constants are given by Eq. 15-14." K_p_1 = exp(-DELTAG_1/(R_u*T_prod_1)) "From EES data" K_P_2 = exp(-DELTAG_2/(R_u*T_prod_2)) "From EES data" "the entahlpy of reaction is estimated from the equilibrium constant K_p by using EQ 15-18 as:" ln(K_P_2/K_P_1)=h_bar_R_Kp/R_u*(1/T_prod_1 - 1/T_prod_2) PercentError = ABS((h_bar_R_enthalpy - h_bar_R_Kp)/h_bar_R_enthalpy)*Convert(, %)


16-95

Percent Error [%]

Tprod [K]

hREnthalpy [kJ/kmol]

hRKp [kJ/kmol]

0.0002739 2000 -251723 -251722 0.0002333 2100 -251920 -251919 0.000198 2200 -252096 -252095 0.0001673 2300 -252254 -252254 0.0001405 2400 -252398 -252398 0.0001173 2500 -252532 -252531 0.00009706 2600 -252657 -252657 0.00007957 2700 -252778 -252777 0.00006448 2800 -252897 -252896 0.00005154 2900 -253017 -253017 0.0000405 3000 -253142 -253142

2000 2200 2400 2600 2800 3000-253250

-252900

-252550

-252200

-251850

-251500

Tprod [k]

h R [

kJ/k

mol

] Enthalpy DataEnthalpy DataKp DataKp Data

DELTATprod = 25 K


16-96

16-107 The KP value of the dissociation process O2 ⇔ 2O at a specified temperature is to be determined using the hR data and KP value at a specified temperature.


Analysis The hR and KP data are related to each other by

⎟⎟⎠

⎞⎜⎜⎝

⎛−≅−⎟⎟

⎠

⎞⎜⎜⎝

⎛−≅

2112

211

2 11lnlnor 11lnTTR

hKK

TTRh

KK

u

RPP

u

R

P

P

The hR of the specified reaction at 2800 K is the amount of energy released as one kmol of O2 dissociates in a steady-flow combustion chamber at a temperature of 2800 K, and can be determined from


Assuming the O2 and O to be ideal gases, we have h = h (T). From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h2800 K

kJ/kmol

O 249,190 6852 59,241

O2 0 8682 98,826

Substituting,

hR = + − − + −=

2 249 190 59 241 6852 1 0 98 826 8682513 014

( , , ) ( , ), kJ / kmol

The KP value at 3000 K can be estimated from the equation above by using this hR value and the KP value at 2600 K which is ln KP1 = -7.521,

⎟⎠⎞

⎜⎝⎛ −

⋅=−−

K 30001

K 26001

KkJ/kmol 314.8kJ/kmol 014,513)521.7(ln 2PK

)357.4ln :28-A (Table 357.4ln 22 −=−= PP KK

or

KP2 = 0.0128


16-97

16-108 It is to be shown that when the three phases of a pure substance are in equilibrium, the specific Gibbs function of each phase is the same.

Analysis The total Gibbs function of the three phase mixture of a pure substance can be expressed as

ggss gmgmgmG ++= ll

where the subscripts s, l, and g indicate solid, liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions, g) constant yields

ggss dmgdmgdmgdG ++= ll

From conservation of mass,

dm dm dm dm dm dms g s g+ + = ⎯ →⎯ = − −l l0

Substituting,

dG g dm dm g dm g dms g g g= − + + +( )l l l

Rearranging,

dG g g dm g g dms g s g= − + −( ) ( )l l

For equilibrium, dG = 0. Also dml and dmg can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. It yields

g g g gs g sl = = and

Combining these two conditions gives the desired result,

g g gs sl = =

mg

ml

ms


16-98

16-109 It is to be shown that when the two phases of a two-component system are in equilibrium, the specific Gibbs function of each phase of each component is the same.

Analysis The total Gibbs function of the two phase mixture can be expressed as

G m g m g m g m gg g g g= + + +( ) ( )l l l l1 1 1 1 2 2 2 2

where the subscripts l and g indicate liquid and gaseous phases. Differentiating by holding the temperature and pressure (thus the Gibbs functions) constant yields

dG g dm g dm g dm g dmg g g g= + + +l l l l1 1 1 1 2 2 2 2

From conservation of mass,

dm dm dm dmg g1 1 2 2= − = −l l and

Substituting,

dG g g dm g g dmg g= − + −( ) ( )l l l l1 1 1 2 2 2

For equilibrium, dG = 0. Also dml1 and dml2 can be varied independently. Thus each term on the right hand side must be zero to satisfy the equilibrium criteria. Then we have

g g g gg gl l1 1 2 2= = and

which is the desired result.

mg1 mg2

ml1 ml2


16-99

16-110 A mixture of CO and O2 contained in a tank is ignited. The final pressure in the tank and the amount of heat transfer are to be determined.

Assumptions 1 The equilibrium composition consists of CO2 and O2. 2 Both the reactants and the products are ideal gases.

Analysis The combustion equation can be written as

CO O CO O2 2 2+ ⎯ →⎯ +3 2 5.

The heat transfer can be determined from

( ) ( )∑∑ −−+−−−+=−RfRPfP PhhhNPhhhNQ vv oooo

out

Both the reactants and the products are assumed to be ideal gases, and thus all the internal energy and enthalpies depend on temperature only, and the vP terms in this equation can be replaced by RuT. It yields

( ) ( )∑∑ −−−−+=−RufRPufP TRhNTRhhhNQ oo

K 829K 050out

since reactants are at the standard reference temperature of 25°C. From the tables,

Substance hfo

kJ/kmol

h298 K

kJ/kmol

h500 K

kJ/kmol

CO -110,530 8669 14,600

O2 0 8682 14,770

CO2 -393,520 9364 17,678

Substituting,

CO kJ/kmol 264,095−=×−−−

×−−×−−++

×−−+−=−

)298314.8530,110(1)298314.80(3

)500314.88682770,140(5.2)500314.89364678,17520,393(1outQ

or

CO kJ/kmol 264,095=outQ

The final pressure in the tank is determined from

atm 2.94=×==⎯→⎯= atm) 2(K 298K 500

45.3

111

222

22

11

2

1 PTNTN

PTRNTRN

PP

u

u

V

V

The equilibrium constant for the reaction CO + O CO12 22 ⇔ is ln KP = 57.62, which is much greater than

7. Therefore, it is not realistic to assume that no CO will be present in equilibrium mixture.

CO2, CO, O2

25°C 2 atm


16-100

16-111 Using Henry’s law, it is to be shown that the dissolved gases in a liquid can be driven off by heating the liquid.

Analysis Henry’s law is expressed as

yP

Hi, liquid sidei, gas side( )

( )0

0=

Henry’s constant H increases with temperature, and thus the fraction of gas i in the liquid yi,liquid side decreases. Therefore, heating a liquid will drive off the dissolved gases in a liquid.


16-101

16-112 A 2-L bottle is filled with carbonated drink that is fully charged (saturated) with CO2 gas. The volume that the CO2 gas would occupy if it is released and stored in a container at room conditions is to be determined.

Assumptions 1 The liquid drink can be treated as water. 2 Both the CO2 gas and the water vapor are ideal gases. 3 The CO2 gas is weakly soluble in water and thus Henry’s law is applicable.

Properties The saturation pressure of water at 17°C is 1.938 kPa (Table A-4). Henry’s constant for CO2 dissolved in water at 17ºC (290 K) is H = 1280 bar (Table 16-2). Molar masses of CO2 and water are 44.01 and 18.015 kg/kmol, respectively (Table A-1). The gas constant of CO2 is 0.1889 kPa.m3/kg.K. Also, 1 bar = 100 kPa.

Analysis In the charging station, the CO2 gas and water vapor mixture above the liquid will form a saturated mixture. Noting that the saturation pressure of water at 17°C is 1.938 kPa, the partial pressure of the CO2 gas is

bar 5.9806=kPa 06.598938.1600C17 @sat vaporside gas ,CO2=−=−=−= °PPPPP

From Henry’s law, the mole fraction of CO2 in the liquid drink is determined to be

0.00467bar 1280bar 9806.5side gas,CO

sideliquid,CO2

2===

H

Py

Then the mole fraction of water in the drink becomes

y ywater, liquid side CO , liquid side2= − = − =1 1 0 00467 0 99533. .

The mass and mole fractions of a mixture are related to each other by

wmm

N MN M

yMMi

i

m

i i

m mi

i

m= = =

where the apparent molar mass of the drink (liquid water - CO2 mixture) is

M y M y M y Mm i i= = +

= × + × =∑ liquid water water CO CO2 2

kg / kmol0 99533 18 015 0 00467 44 01 1814. . . . .

Then the mass fraction of dissolved CO2 in liquid drink becomes

w yMMm

CO , liquid side CO , liquid sideCO

2 2

2 0.0113= = =( ) ...

0 0 0046744 011814

Therefore, the mass of dissolved CO2 in a 2 L ≈ 2 kg drink is

m w mmCO CO2 2 kg) 0.0226 kg= = =0 0113 2. (

Then the volume occupied by this CO2 at the room conditions of 20°C and 100 kPa becomes

L 12.5m 0.0125 3 ==⋅⋅

==kPa 100

K) 293)(Kkg/mkPa kg)(0.1889 0226.0( 3

PmRT

V

Discussion Note that the amount of dissolved CO2 in a 2-L pressurized drink is large enough to fill 6 such bottles at room temperature and pressure. Also, we could simplify the calculations by assuming the molar mass of carbonated drink to be the same as that of water, and take it to be 18 kg/kmol because of the very low mole fraction of CO2 in the drink.


16-102

16-113 EES Ethyl alcohol C2H5OH (gas) is burned in a steady-flow adiabatic combustion chamber with 40 percent excess air. The adiabatic flame temperature of the products is to be determined and the adiabatic flame temperature as a function of the percent excess air is to be plotted.

Analysis The complete combustion reaction in this case can be written as

[ ] 22th2222th52 N O ))((OH 3CO 23.76NO)1((gas) OHHC faExaEx +++⎯→⎯+++

where ath is the stoichiometric coefficient for air. The oxygen balance gives

2))((13222)1(1 thth ×+×+×=×++ aExaEx

The reaction equation with products in equilibrium is

[ ] 222222th52 N O OH CO CO 3.76NO)1((gas) OHHC fedbaaEx ++++⎯→⎯+++




Oxygen balance: 222)1(1 th ×+++×=×++ edbaaEx

Nitrogen balance: faEx =×+ 76.3)1( th

Solving the above equations, we find the coefficients to be

Ex = 0.4, ath = 3, a = 1.995, b = 0.004938, d = 3, e = 1.202, f = 15.79

Then, we write the balanced reaction equation as

[ ] 22222252 N 79.15O 202.1OH 3CO 004938.0CO 995.13.76NO2.4(gas) OHHC ++++⎯→⎯++


99.2179.15202.13004938.0995.1tot =++++=N


22 O5.0COCO +⎯→←



p uu= = −−Δ Δ*( )/ ln *( ) / or

where

)()()()(* prodCO2CO2prodO2O2prodCOCO TgTgTgTG ∗∗∗ −+=Δ ννν


CO2prodprodCO2

O2prodprodO2

COprodprodCO

)()(

)()(

)()(

sThTg

sThTg

sThTg

−=

−=

−=

∗

∗

∗


0005787.099.21

1995.1

)202.1)(004938.0( 5.05.015.01

tot

5.0=⎟

⎠⎞

⎜⎝⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

−+

NP

abeK p


16-103

A steady flow energy balance gives

PR HH =

where

kJ/kmol 310,235.79(0)15(0)2.4kJ/kmol) 310,235(

79.152.4 CN2@25CO2@25Cfuel@25

−=++−=

++= °°°hhhH o

fR

prodprodprodprodprod N2@O2@H2O@CO@CO2@ 79.15202.13004938.0995.1 TTTTTP hhhhhH ++++=

Solving the energy balance equation using EES, we obtain the adiabatic flame temperature to be

K 1907=prodT

The copy of entire EES solution including parametric studies is given next:

"The product temperature isT_prod" "The reactant temperature is:" T_reac= 25+273.15 "[K]" "For adiabatic combustion of 1 kmol of fuel: " Q_out = 0 "[kJ]" PercentEx = 40 "Percent excess air" Ex = PercentEx/100 "EX = % Excess air/100" P_prod =101.3"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=2 CO2 + 3 H2O +Ex*A_th O2 + f N2" "Oxygen Balance for complete combustion:" 1 + (1+Ex)*A_th*2=2*2+3*1 + Ex*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C2H5OH(gas)+ (1+Ex)*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 2=a + b "Hydrogen Balance:" 6=2*d "Oxygen Balance:" 1 + (1+Ex)*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" (1+Ex)*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod )) P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15."


16-104

"K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P h_bar_f_C2H5OHgas=-235310 "[kJ/kmol]" H_R=1*(h_bar_f_C2H5OHgas ) +(1+Ex)*A_th*ENTHALPY(O2,T=T_reac)+(1+Ex)*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"

a ath b d e f PercentEx [%]

Tprod [K]

1.922 3 0.07809 3 0.339 12.41 10 2191 1.97 3 0.03017 3 0.6151 13.54 20 2093

1.988 3 0.01201 3 0.906 14.66 30 1996 1.995 3 0.004933 3 1.202 15.79 40 1907 1.998 3 0.002089 3 1.501 16.92 50 1826 1.999 3 0.0009089 3 1.8 18.05 60 1752

2 3 0.000405 3 2.1 19.18 70 1685 2 3 0.0001843 3 2.4 20.3 80 1625 2 3 0.0000855 3 2.7 21.43 90 1569 2 3 0.00004036 3 3 22.56 100 1518

10 20 30 40 50 60 70 80 90 100

1500

1600

1700

1800

1900

2000

2100

2200

PercentEx

T pro

d (K

)


16-105

16-114 EES The percent theoretical air required for the combustion of octane such that the volume fraction of CO in the products is less than 0.1% and the heat transfer are to be determined. Also, the percent theoretical air required for 0.1% CO in the products as a function of product pressure is to be plotted.

Analysis The complete combustion reaction equation for excess air is

[ ] 22thth2222thth188 N O )1(OH 9CO 83.76NOHC faPaP +−++⎯→⎯++

The oxygen balance is

2)1(19282 thththth ×−+×+×=× aPaP

The reaction equation for excess air and products in equilibrium is

[ ] 222222thth188 N O OH CO CO 3.76NOHC fedbaaP ++++⎯→⎯++

The coefficients are to be determined from the mass balances



Oxygen balance: 222thth ×+++×=× edbaaP

Nitrogen balance: faP =× 76.3thth

Volume fraction of CO must be less than 0.1%. That is,

001.0tot

CO =++++

==fedba

bN

by


22 O5.0COCO +⎯→←

The Kp value of a reaction at a specified temperature can be determined from the Gibbs function data:

kJ/kmol 121,920)00.309)(2000()128,302()()(

kJ/kmol 876,477)53.268)(2000()193,59()()(

kJ/kmol 781,570)48.258)(2000()826,53()()(

CO2prodprodCO2

O2prodprodO2

COprodprodCO

−=−−=−=

−=−=−=

−=−−=−=

∗

∗

∗

sThTg

sThTg

sThTg

The enthalpies at 2000 K and entropies at 2000 K and 101.3 kPa are obtained from EES. Substituting,

kJ/kmol 402,110)121,920()876,477(5.0)781,570(1

)()()()(* prodCO2CO2prodO2O2prodCOCOprod

=−−−+−=

−+=Δ ∗∗∗ TgTgTgTG ννν

001308.0)2000)(314.8(

402,110exp)(*

expprod

prod =⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛ Δ−=

TRTG

Ku

p


15.01

prod5.015.01

tot

5.0 3.101/ −+−+

⎟⎟⎠

⎞⎜⎜⎝

⎛

++++=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

fedbaP

abe

NP

abeK p

The steady flow energy balance gives

PR HQH += out

where


16-106

)115,56()193,59()171,169()826,53()128,302(

kJ/kmol 459,208)0)(76.3()0()459,208(

)76.3(1

K 2000 @ N2K 2000 @ O2K 2000 @ H2OK 2000 @ COK 2000 @ CO2

thththth

K 298 @ N2ththK 298 @ O2ththK 298 @ C8H18

fedba

hfhehdhbhaH

aPaP

haPhaPhH

P

R

++−+−+−=

++++=

−=×++−=

×++=

The enthalpies are obtained from EES. Solving all the equations simultaneously using EES, we obtain

188HC kJ/kmol 995,500102.4%

==×=×=

=======

out

th

thth

100024.1100PercentTh11.48 ,3289.0 ,9 ,06544.0 ,935.7 ,5.12 ,024.1

QP

fedbaaP

The copy of entire EES solution including parametric studies is given next:

"The product temperature is:" T_prod = 2000 "[K]" "The reactant temperature is:" T_reac= 25+273 "[K]" "PercentTH is Percent theoretical air" Pth= PercentTh/100 "Pth = % theoretical air/100" P_prod = 5 "[atm]" *convert(atm,kPa)"[kPa]" R_u=8.314 "[kJ/kmol-K]" "The complete combustion reaction equation for excess air is:" "C8H18+ Pth*A_th (O2 +3.76N2)=8 CO2 + 9 H2O +(Pth-1)*A_th O2 + f N2" "Oxygen Balance for complete combustion:" Pth*A_th*2=8*2+9*1 + (Pth-1)*A_th*2 "The reaction equation for excess air and products in equilibrium is:" "C8H18+ Pth*A_th (O2 +3.76N2)=a CO2 + b CO+ d H2O + e O2 + f N2" "Carbon Balance:" 8=a + b "Hydrogen Balance:" 18=2*d "Oxygen Balance:" Pth*A_th*2=a*2+b + d + e*2 "Nitrogen Balance:" Pth*A_th*3.76 = f N_tot =a +b + d + e + f "Total kilomoles of products at equilibrium" "The volume faction of CO in the products is to be less than 0.1%. For ideal gas mixtures volume fractions equal mole fractions." "The mole fraction of CO in the product gases is:" y_CO = 0.001 y_CO = b/N_tot "The assumed equilibrium reaction is CO2=CO+0.5O2" "The following equations provide the specific Gibbs function (g=h-Ts) for each component in the product gases as a function of its temperature, T_prod, at 1 atm pressure, 101.3 kPa" g_CO2=Enthalpy(CO2,T=T_prod )-T_prod *Entropy(CO2,T=T_prod ,P=101.3) g_CO=Enthalpy(CO,T=T_prod )-T_prod *Entropy(CO,T=T_prod ,P=101.3) g_O2=Enthalpy(O2,T=T_prod )-T_prod *Entropy(O2,T=T_prod ,P=101.3) "The standard-state Gibbs function is" DELTAG =1*g_CO+0.5*g_O2-1*g_CO2 "The equilibrium constant is given by Eq. 15-14." K_P = exp(-DELTAG /(R_u*T_prod ))


16-107

P=P_prod /101.3"atm" "The equilibrium constant is also given by Eq. 15-15." "K_ P = (P/N_tot)^(1+0.5-1)*(b^1*e^0.5)/(a^1)" sqrt(P/N_tot )*b *sqrt(e )=K_P *a "The steady-flow energy balance is:" H_R = Q_out+H_P H_R=1*ENTHALPY(C8H18,T=T_reac)+Pth*A_th*ENTHALPY(O2,T=T_reac)+Pth*A_th*3.76*ENTHALPY(N2,T=T_reac) "[kJ/kmol]" H_P=a*ENTHALPY(CO2,T=T_prod)+b*ENTHALPY(CO,T=T_prod)+d*ENTHALPY(H2O,T=T_prod) +e*ENTHALPY(O2,T=T_prod)+f*ENTHALPY(N2,T=T_prod) "[kJ/kmol]"

Pprod [kPa] PercentTh [%] 100 112 300 104.1 500 102.4 700 101.7 900 101.2

1100 101 1300 100.8 1500 100.6 1700 100.5 1900 100.5 2100 100.4 2300 100.3

0 500 1000 1500 2000 2500100

102

104

106

108

110

112

Pprod [kPa]

Perc

entT

h %


16-108

Fundamentals of Engineering (FE) Exam Problems 16-115 If the equilibrium constant for the reaction H2 + ½O2 → H2O is K, the equilibrium constant for the reaction 2H2O → 2H2 + O2 at the same temperature is (a) 1/K (b) 1/(2K) (c) 2K (d) K2 (e) 1/K2 Answer (e) 1/K2 16-116 If the equilibrium constant for the reaction CO + ½O2 → CO2 is K, the equilibrium constant for the reaction CO2 + 3N2 → CO + ½O2 + 3N2 at the same temperature is (a) 1/K (b) 1/(K + 3) (c) 4K (d) K (e) 1/K2 Answer (a) 1/K 16-117 The equilibrium constant for the reaction H2 + ½O2 → H2O at 1 atm and 1500°C is given to be K. Of the reactions given below, all at 1500°C, the reaction that has a different equilibrium constant is (a) H2 + ½O2 → H2O at 5 atm, (b) 2H2 + O2 → 2H2O at 1 atm, (c) H2 + O2 → H2O+ ½O2 at 2 atm, (d) H2 + ½O2 + 3N2 → H2O+ 3N2 at 5 atm, (e) H2 + ½O2 + 3N2 → H2O+ 3N2 at 1 atm, Answer (b) 2H2 + O2 → 2H2O at 1 atm, 16-118 Of the reactions given below, the reaction whose equilibrium composition at a specified temperature is not affected by pressure is (a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) all of the above. Answer (c) N2 + O2 → 2NO


16-109

16-119 Of the reactions given below, the reaction whose number of moles of products increases by the addition of inert gases into the reaction chamber at constant pressure and temperature is (a) H2 + ½O2 → H2O (b) CO + ½O2 → CO2 (c) N2 + O2 → 2NO (d) N2 → 2N (e) none of the above. Answer (d) N2 → 2N 16-120 Moist air is heated to a very high temperature. If the equilibrium composition consists of H2O, O2, N2, OH, H2, and NO, the number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (c) 3 16-121 Propane C3H8 is burned with air, and the combustion products consist of CO2, CO, H2O, O2, N2, OH, H2, and NO. The number of equilibrium constant relations needed to determine the equilibrium composition of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (d) 4 16-122 Consider a gas mixture that consists of three components. The number of independent variables that need to be specified to fix the state of the mixture is (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 Answer (d) 4


16-110

16-123 The value of Henry’s constant for CO2 gas dissolved in water at 290 K is 12.8 MPa. Consider water exposed to air at 100 kPa that contains 3 percent CO2 by volume. Under phase equilibrium conditions, the mole fraction of CO2 gas dissolved in water at 290 K is (a) 2.3×10-4 (b) 3.0×10-4 (c) 0.80×10-4 (d) 2.2×10-4 (e) 5.6×10-4 Answer (a) 2.3×10-4 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). H=12.8 "MPa" P=0.1 "MPa" y_CO2_air=0.03 P_CO2_air=y_CO2_air*P y_CO2_liquid=P_CO2_air/H "Some Wrong Solutions with Common Mistakes:" W1_yCO2=P_CO2_air*H "Multiplying by H instead of dividing by it" W2_yCO2=P_CO2_air "Taking partial pressure in air" 16-124 The solubility of nitrogen gas in rubber at 25°C is 0.00156 kmol/m3⋅bar. When phase equilibrium is established, the density of nitrogen in a rubber piece placed in a nitrogen gas chamber at 800 kPa is (a) 0.012 kg/m3 (b) 0.35 kg/m3 (c) 0.42 kg/m3 (d) 0.56 kg/m3 (e) 0.078 kg/m3 Answer (b) 0.35 kg/m3 Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T=25 "C" S=0.00156 "kmol/bar.m^3" MM_N2=28 "kg/kmol" S_mass=S*MM_N2 "kg/bar.m^3" P_N2=8 "bar" rho_solid=S_mass*P_N2 "Some Wrong Solutions with Common Mistakes:" W1_density=S*P_N2 "Using solubility per kmol" 16-125 … 16-128 Design and Essay Problems

Solution manual of Thermodynamics-Ch.16

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